pls help! I'll give u 10 points and the brainlest answer!!!
Answer:
371km
Explanation:
the lower the more fuel there is
A. Express each Fraction in Decimal form. Round off your answers into
thousandths place.
1. 3/4
Answer:
0.75
Explanation:
3/4=0.75 so therefore its 0.75
A graduated cylinder is filled to an initial volume of 25mL. A solid object is dropped into the graduated cylinder. The final volume of the graduated cylinder is 60mL. The mass of the object is 140g. What is the density of the solid object?
Answer:
The answer is 4 g/mLExplanation:
The density of a substance can be found by using the formula
[tex]density = \frac{mass}{volume} \\ [/tex]
volume = final volume of water - initial volume of water
volume = 60 - 25 = 35 mL
From the question we have
[tex]density = \frac{140}{35} \\ [/tex]
We have the final answer as
4 g/mLHope this helps you
10. Which of the following is NOT correctly matched?
A. Salt: Na and C: Element
B. Water: H and O: Corapound
C. Carbon dioxside: C and O: Compound
D. Magnesium chloride: Mg and C: Compound
Answer:
A. because salt : Na and Cl, not C
Which of the following are decomposers?
O A. mice
O B. snakes
O C. fungi
OD. spiders
ASAP
A scientist plans to find out the cause of decrease in bird population in a location. He interviews 50 people who work in the local oil mine and 45 of them believe that the mine does not affect the birds. The scientist concludes that 90 percent of people in the location believe that the local oil mine does not cause decrease in bird population. Why is the scientist's conclusion most likely unreliable? (2 points)
Group of answer choices
a. the results of the investigation cannot be repeated
b. the source of information could be biased
c. bird population cannot be measured accurately
d. the investigation is done by only one person
Answer:
Explanation:
6. Clownfish, sea anemone, coral, sea urchins, sponges and parrot fish are 5 points
just a few examples of the large variety of organisms that can be found in
a coral reef. The coral reef and its inhabitants (residents) are suffering due
to several factors such as pollution, coastal development, sedimentation
and destructive fishing practices. What would be the most likely result if
some of the species in the coral reef are destroyed? *
Answer:
The orange-and-white clownfish you know from Hollywood is only one of 30 species of clownfish that live in coral reefs around the world; small in size and big in personality, these tiny comedians have one of the most unique lifestyles in the reef. The orange-and-white clownfish you know from Hollywood is only one of 30 species of clownfish that live in coral reefs around the world; small in size and big in personality, these tiny comedians have one of the most unique lifestyles in the reef.
Explanation:
point being, it would be absolutely terrible
Coral reef are the biggest biodiversity in the world. Therefore, destruction of coral reefs definitely affect the aquatic livings. As well as it leads to erosion of seashore lines.
What are coral reefs?Corals reefs are the habitat of 25% of sea organisms. It means there are well over a million species that rely on coral reefs for their existence. The reef offers these organisms the vital food, habitat, and breeding grounds required for the survival of their species.
The marine biodiversity would suffer greatly if their homes vanished. And many fish, turtles, and other species would vanish in a domino-like fashion.
Coral reefs are threatened by a number of things, such as local threats like overfishing, destructive fishing methods, coastal development, pollution, and careless tourism, as well as global effects of climate change like warming seas and rising CO₂ levels in the water. Natural phenomena like hurricanes, El Nio, and diseases are also a threat to oral reefs.
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The indicator dilution method is a technique used to determine flow rates of fluids in channels for which devices like rotameters and orifice meters cannot be used (e.g., rivers, blood vessels, and large- diameter pipelines). A stream of an easily measured substance (the tracer) is injected into the channel at a known rate and the tracer concentration is measured at a point far enough downstream of the injection point for the tracer to be completely mixed with the flowing fluid. The larger the flow rate of the fluid, the lower the tracer concentration at the measurement point. A gas stream that contains 1.50 mole% CO2 flows through a pipeline, Twenty (20.0) kilograms of CO2 per minute is injected into the line. A sample of the gas is drawn from a point in the line 150 meters downstream of the injection point and found to contain 2.3 mole% CO2.(a) Estimate the gas flow rate (k mol/min) upstream of the injection point.(b) Eighteen seconds elapses from the instant the additional CO2 is first injected to the time the CO2 concentration at the measurement point begins to rise. Assuming that the tracer travels at the average velocity of the gas in the pipeline (i.e., neglecting diffusion of CO2), estimate the average velocity (m/s), if the molar gas density is 0.123k mol/m3 what is the pipe diameter?
Answer:
[a]. 55.5kmol/min.
[b]. The average velocity = 8.3m/s.
pipe diameter = 1.0954m.
Explanation:
So, from the question above we are given the following data or information or parameters which are going to help us in solving the above question;
=> ''A gas stream that contains 1.50 mole% CO2 flows through a pipeline''
=> ''Twenty (20.0) kilograms of CO2 per minute is injected into the line.''
=> '' . A sample of the gas is drawn from a point in the line 150 meters downstream of the injection point and found to contain 2.3 mole% CO2''
=> ''molar gas density is 0.123k mol/m3''
[a].The gas flow rate (k mol/min) upstream of the injection point can be calculated as follows.
- Determine the mass balance= j₁ +[ 20 /.123 × 44] = j₃. Thus, the flow gas rate is given as;
w₁ × 1.50 mole% CO2 + 3.6954= 2.3 mole% CO2 × w₃.
-3.610 = - 0.008w₁.
w₁ = 451.3 m³/min.
Thus, the flow rate = 0.123 × 451.3 = 55.51 kmol/min.
[b]. The average velocity = 150/ 18 seconds = 8.3 m/s.
[c]. Pipe Diameter can be calculated as follows;
0.123 × 8.3 × [3.12 × (pipe diameter)² / 4] = 55.51/60.
[1.0209 × 3.12]/4 × [pipe diameter]² = 0.9252.
0.8014065 × [pipe diameter]² = 0.9252.
[pipe diameter]² = 0.9252/ .8014065.
[pipe diameter]² = 1.15443.
[pipe diameter] = √1.15443.
[pipe diameter] = 1.2 m² = 1.0954m.
A) The gas flow rate (k.mol/min) upstream of the injection point is;
55.51 k.mol/min
B) The average velocity and the diameter of the pipe are respectively;
8.33 m/s and 1.072 m
Gas Flow rate in thermodynamics
Due to the fact that there is no reaction or accumulation in the pipeline, we can say that the input of CO₂ is equal to it's output. Thus;
Upstream + Injection = Downstream
A) If we assume that x is the flow rate of gas (kmol/min), then there is the we can write the balance as:
0.015x + (20/44) = 0.023(x + 20/44)
Expanding gives us;
0.015x + 0.4545 = 0.023x + 0.01045
Solving for x gives;
x = 55.51 kmol/min
B) Formula for velocity is;
v = distance/time
Thus, since the length of the pipe is 150 m/s and time is 18 seconds, then velocity is;
v = 150/18
v = 8.33 m/s
Now, volumetric flow rate is calculated from;
Q = q' * ρ
where;
q' is gas flow rate
ρ is gas density
Thus;
Q = 55.51 * 0.123
Q = 451.3 m³/min = 7.522 m³/s
We can also express average velocity as;
v = Q/A
where A is area = πD²/4
Thus;
8.33 = 7.522/(πD²/4)
Making D the subject of the formula gives;
D = √((4 * 7.522)/(8.33π))
D = 1.072 m
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If a sample of CO2 contains 48 g of oxygen, how many moles of CO2 are there?
Answer:
0.75 moles of CO2
Explanation:
molar mass of CO2⇒ 44.01 g/mol
molar mass Oxygen ⇒ 31.998 g/mol
divide the mass given by molar mass of oxygen so
48÷31.998= 1.50009376 moles of O
then you use the ration of oxygen to carbon to find the moles of CO2 which is one mole of CO2 for 2 moles of oxygen
1.50009376 moles of O ×[tex]\frac{1}{2}[/tex] = 0.75004688
with sig figs
0.75 moles of CO2
Identify the change of state occurring in each situation. your friend's perfume producing an aroma that fills the room making ice cream from milk in an ice cream maker formation of frost on a cold windshield overnight wax dripping from a candle naphthalene solid (found in moth balls) producing fumes in the closet to deter moths your breath fogging up on your car windshield on a cold winter day
Answer:
The solution to this question can be defined as follows:
Explanation:
[tex]W_{ax}[/tex] Naphthalene falls off from candle something, strong fumes-sublimation produces. On a rainy day in winter, the breath fogging up on your cor's windshield. Brost formation on an overnight deposition on a frozen windshield. Making milk ice cream freezing the perfuming scent that fills the creation of the room.the de broglie equation predicts that the wavelength (in m) of a proton moving at 1000 m/s is what?
(h=6.63 x 10^-34J•s; mass od a photon = 1.673 x 10^-24
Explanation:
from
de Broglie wavelength= h/ mv
substitute the values
joules= mass(kg) × acceleration (m/s^2) × distance(m)= kgm^2/s^2.
sorry I don't have my calc with me
According to De Broglie's Hypothesis, matter has both particle and wave properties. The wavelength of De Broglie is given as λ = h p,
wavelength (in m) of a proton = 11.09 x[tex]10^ {-55[/tex] m.
What is de Broglie statement?According to De Broglie's Hypothesis, matter has both particle and wave properties. The wavelength of De Broglie is given as λ = h p, where represents the particle momentum, and can be written as: λ= h m v.The de Broglie hypothesis demonstrated that wave-particle duality was a fundamental principle shared by both radiation and matter, rather than an aberrant behavior of light.De Broglie's equation does not apply to larger particles; rather, it only applies to microscopic particles such as electrons, protons, and neutrons. De Broglie's equation is only valid for moving microscopic particles.λ= h m v.
= 6.63 x [tex]10^{-34}[/tex] x 1.673 x [tex]10^{-24[/tex]x 1000
= 11.09 x[tex]10^ {-55[/tex] m
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H3C - CH2 - CEC - CH3
Compound name
Answer:
2-pentene
hope this helps :)
Explanation:
1.547 grams of hydrated MgSO4 is heated in a crucible. After heating, 0.7554 grams of anhydrous MgSO4 remains in the crucible. How many waters of hydration were attached to the MgSO4
Answer:
7
Explanation:
Let x represent the number of moles of water in the hydrated salt i.e MgSO₄.xH₂O
The following data were obtained from the question:
Mass of MgSO₄.xH₂O = 1.547 g
Mass of anhydrous MgSO₄ = 0.7554 g
Mole of H₂O = x =?
Next, we shall determine the mass of water, H₂O in the hydrated salt, MgSO₄.xH₂O. This can be obtained as follow:
Mass of MgSO₄.xH₂O = 1.547 g
Mass of anhydrous MgSO₄ = 0.7554 g
Mass of H₂O =?
Mass of H₂O = (Mass of MgSO₄.xH₂O) – (Mass of anhydrous MgSO₄)
Mass of H₂O = 1.547 – 0.7554
Mass of H₂O = 0.7916 g
Finally, we shall determine the value of the x as illustrated below:
Mass of MgSO₄.xH₂O = 1.547 g
Molar mass of MgSO₄.xH₂O = 24 + 32 + (16×4) + x[(2×1) + 16]
= 24 + 32 + 64 + x(2 + 16)
= 120 + 18x
Mass of H₂O = 0.7916 g
Molar mass of xH₂O = 18x
Molar Mass of xH₂O/ Molar mass of MgSO₄.xH₂O = mass of xH₂O /Mass of MgSO₄.xH₂O
18x/ 120 + 18x = 0.7916/1.547
Cross multiply
0.7916 (120 + 18x) = 18x × 1.547
94.992 + 14.2488x = 27.846x
Collect like terms
94.992 = 27.846x – 14.2488x
94.992 = 13.5972x
Divide both side by 13.5972
x = 94.992 / 13.5972
x = 7
Thus, the formula for the hydrated salt, MgSO₄.xH₂O is MgSO₄.7H₂O
Number of moles of water, H₂O in the hydrated salt MgSO₄.7H₂O is 7.
The number of moles of attached water molecules is 7.
Mass of hydrated MgSO4 = 1.547 grams
Mass of anhydrous MgSO4 = 0.7554 grams
Number of moles of hydrated MgSO4 = 1.547 grams/120 + 18x
Number of moles of anhydrous MgSO4 = 0.7554 grams /120
Number of moles of anhydrous salt = Number of moles of hydrated salt
0.7554 grams /120 = 1.547 grams/120 + 18x
0.7554(120 + 18x) = 1.547 × 120
90.6 + 13.6x = 185.6
185.6 - 90.6 /13.6 = x
x = 7
The number of moles of attached water molecules is 7.
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How many F atoms are present in 3.16 moles of BaF2? show work
Answer:
3.80 × 10²⁴ atoms of F
Explanation:
Step 1: Given data
Moles of barium fluoride (BaF₂): 3.16 mol
Step 2: Calculate the moles of fluorine in 3.16 moles of barium fluoride
The molar ratio of BaF₂ to F is 1:2. The moles of F in 3.16 moles of BaF₂ are 2/1 × 3.16 mol = 6.32 mol.
Step 3: Calculate the atoms of F in 6.32 moles of F
We will use Avogadro's number: there are 6.02 × 10²³ atoms of F in 1 mole of atoms of F.
6.32 mol × (6.02 × 10²³ atom/1 mol) = 3.80 × 10²⁴ atom
A 50.0 g sample of an unknown substance, initially at 20.2 °C, was heated with 1.55 kJ of energy. The final temperature of the substance was 125.0 °C. Determine the specific heat of this substance.
Answer:
0.296j/g⁰c
Explanation:
we have the following information from this question before us.
mass iv substance = 50grams
we have initial temperature ti = 20.2⁰c
final temperature = 125⁰c
the energy that was provided = 155kj
we proceed with this formula
energy = mcΔT
1.55x10³ = 50 x c x (125-20.2)
1.55x10³ = c x 50gm x 104.8k
we divide through to get c
c = 1.55x10³/50g x 104.8
c = 0.296J/g⁰c
that is the specific heat of this substance.
thank you!
Sketch the titration curve for the titration of a generic weak base B with a strong acid. The titration reaction is
B+H+⇌BH+
On the curve indicate the points that correspond to the following:
a) the equivalence point.
b) the region with maximum buffering) the point where pH = pKbd) the region where pH depends only on [B]e) the region where pH depends only on [BH+]f) the region where pH depends only on the amount of excess strong acid
Answer:
Kindly check the attached picture.
Explanation:
Titration is one of the useful concept or practical in chemistry which is used in the determination of the concentration of a particular chemical specie. Without mincing words, let us dive straight into the solution to the question;
(a) Determining the equivalence point using the graph.
=> The equivalence point is the point at which there has been been "enough" addition of the acid to the base to give us the congugate acid. In the diagram showing the curve, it is the point labelled (a).
(b)The next thing is to determine the region with maximum buffering.
=> The position with the maximum buffering ranges from the point labelled (b) in the graph and it is the point when the acid and the base has been added. But the point with the maximum buffering is labeled as point (c).
(c). Determination of the point where pH = pKb.
=> The Point in which the pH = pKb is the point a
(d). The region where pH depends only on [B].
=> The starting point in the point in which the pH is dependent on the concentration of the base, [B].
(e). The region where pH depends only on [BH+].
=> This is the same point with point labelled as (a). Thus, point (a) and (e) are the same.
(f). The region where pH depends only on the amount of excess strong acid is the point labelled as (f) which is known as the post equivalence point.
The density for potassium is 0.856 g/cm3. What would be the mass of a 35 cm3 piece of potassium?
Answer:
30 grams
Explanation:
density = mass / volume => mass = density x volume
mass = 0.856 g/cm³ x 35 cm³ = 29.96 grams ≅ 30 grams 2 sig. figs.
Li atoms in 3.7 moles
Answer:
We know that there are 6.022 * 10²³ atoms in 1 mole of any element
So, that means that there are 6.022 * 10²³ atoms in 1 mole of Li (Lithium)
Atoms in 3.7 Moles:
atoms in one mole = 6.022 * 10²³
atoms in 3.7 moles = 6.022 * 10²³ * 3.7
atoms in 3.7 moles = 22.28 * 10²³
atoms in 3.7 moles = 2.228 * 10²⁴
I need help pls... look at photo
Answer:
LA PRIMERA ES N
Explanation:
List the following elements in order of INcreasing number of valence electrons: C, Cl, As, Na, He
Answer:
He, Cl, As, C, Na
Explanation:
^^
using this chemical might discolour equipment and surfaces
Answer: do not know how to answer this
Explanation:
Which of the following is an Elementary compound?
A. CO2
B. N2
C. SO2
D. H2S
heeeeeeeeeelp please please please
Answer:
Explanation:
In my opinion the answer should be SO2
Answer:
a should be answer i think.
What is the reason behind an atom being neutral?
smallest to largest: mountain system, mountain range, mountain belt, mountain.
Answer:
mountain, mountain range, mountain system, mountain belt
Explanation:
edge 2021
1) A hanglider flies with a horizontal velocity of 13 m/s when his wallet falls out. If the hanglider is 1450 m above the earth when the wallet falls out, find the horizontal distance the wallet travels before hitting the ground.
Answer:
The wallet is falling from a height of 1450 m
We are given that:
downward acceleration (a) = 10 m/s/s
downward displacement (s) = 1450 m
downward initial velocity (u) = 0 m/s (Since the wallet was dropped, it had no initial velocity)
horizontal initial velocity (v) = 13 m/s
Since there is no external force being applied on the wallet except the force of gravity, the wallet will keep moving at a constant velocity of 13 m/s
To calculate the horizontal distance travelled by the wallet, we need to know how long the wallet was airborne
To calculate how long the wallet was airborne, we have to find out that how long it took the wallet to hit the ground
Time taken for the wallet to hit the ground:
From the second equation of motion:
s = ut + 1/2 at² ------------(for vertical motion of the wallet)
1450 = (0)(t) + 1/2 (10)(t)²
1450 = 5t²
t² = 290
t = 17 (approx)
Time taken for the wallet to hit the ground = 17 seconds
Horizontal distance travelled by the wallet:
Since the wallet hits the ground after 17 seconds, it will move horizontally at a constant velocity of 13 m/s for 17 seconds
horizontal acceleration = 0 m/s/s
From the second equation of motion:
s = ut + 1/2at²
s = ut + 1/2 (0)t²
s = ut
here, u is the horizontal initial velocity of the wallet and the time taken by the wallet to hit the ground is 't'
s = (13)(17)
s = 221 m
Hence, the horizontal distance travelled by the wallet while falling from a height of 1450 m is 221 m
A boiling pot of soup with a lid on it to keep in the energy is what type of system?
open
closed
isolated
none of the above
Answer: In a closed system, only energy can be exchanged between the system and the surroundings. In the same example of boiling a pot of water, putting the lid on the system changes it from open to closed. This is because when the lid is on the pot, matter can no longer escape.
Explanation:
Answer:
I think its closed
Explanation:
Sorry if its wrong
Which of the following favors the solubility of an ionic solid in a liquid solvent? Group of answer choices a large magnitude of the solvation energy of the ions a small magnitude of the lattice energy of the solute a large polarity of the solvent all of the above none of the above
Answer:
all of the above
Explanation:
The solubility of an ionic solid in a liquid solvent largely depends on three factors:
i) The solvation energy
ii) The lattice energy of the solid
iii) The polarity of the of the solvent
A large solvation energy and a small magnitude of lattice energy means that the ionic solute will dissolve very easily in the solvent.
Secondly, the solvent must have a large polarity, this is why water serves as a good solvent for most ionic solutes.
The pH of a solution of 19.5 g of malonic acid in 0.250 L is 1.47.The pH of a 0.300 M Solution of sodium hydrogen malonate is 4.26. What are the values of Ka1and Ka2
Answer:
[tex]Ka_1=1.61x10^{-3}[/tex]
[tex]Ka_2=1.01x10^{-8}[/tex]
Explanation:
Hello.
In this case, since the stepwise dissociation of malonic acid which is a diprotic acid that we are going to symbolize by H₂A, is:
[tex]H_2A\rightleftharpoons H^++HA^-;Ka_1\\\\HA^-\rightleftharpoons H^++A^-;Ka_2[/tex]
The first ionization has the following equilibrium expression:
[tex]Ka_1=\frac{[H^+][HA^-]}{[H_2A]}[/tex]
Whereas the concentration of H⁺ equals the concentration of HA⁻ and is computed via the pH:
[tex][H^+]=[HA^-]=10^{-pH}=10^{-1.47}=0.0339M[/tex]
Next, we compute the molarity of the 19.5 g of malonic acid (molar mass = 104.06 g/mol) as shown below:
[tex][H_2A]=\frac{19.5g/(104.06 g/mol)}{0.250L}=0.750M[/tex]
Thus, Ka1 turns out:
[tex]Ka_1=\frac{(0.0339)(0.0339)}{0.750-0.0339}=1.61x10^{-3}[/tex]
Now, for the second ionization, since the 0.300-M sodium hydrogen malonate is the source of HA⁻, and the pH is 4.26, we can compute the concentration of both H⁺ and A⁻² again by considering the pH:
[tex][H^+]=[A^-^2]=10^{-4.26}=5.50x10^{-5}M[/tex]
Therefore Ka2 turns out:
[tex]Ka_2=\frac{[H^+][A^{-2}]}{[HA^-]}=\frac{(5.50x10^{-5})(5.50x10^{-5})}{0.300-(5.50x10^{-5})}\\ \\Ka_2=1.01x10^{-8}[/tex]
Best regards!
Facilitated diffusion is a form of passive transport and involves the movement of molecules (like glucose) that are too large to just go through the cell membrane without help. These larger molecules move through the cell membrane from High concentration to Low concentration with the help of _____________________________.
Answer: Membrane Protiens
Explanation:
In facilitated diffusion, molecules diffuse across the plasma membrane with assistance from membrane proteins, such as channels and carriers. A concentration gradient exists for these molecules, so they have the potential to diffuse into (or out of) the cell by moving down it.
A student dissolves of sodium hydroxide in of water in a well-insulated open cup. He that observes the temperature of the water rise from to over the course of minutes. Use this data, and any information you need from the ALEKS Data resource,
NaOH(s) rightarrow Na+(aq) + OH-(aq)
You can make any reasonable assumptions about the physical properties of the solution. Note for advanced students: it's possible the student did not do the experiment carefully, and the values you calculate may not be the same as the known and published values for this reaction.
1. Is this reaction exothermic, endothermic, or neither.
2. If you said the reaction was exothermic or endothermic, calculate the amount of neat that was released or absorbed by the reaction in this case.
3. Calculate the reaction enthalpy AW per mole of NaOH.
Answer:
This question is incomplete
Explanation:
This question is incomplete but the answer to (1) and steps involved in calculating (2) and (3) will be discussed in the attachment
1. The reaction is an exothermic reaction because heat was given off. Exothermic reaction is a reaction in which heat is released from a reaction into the surroundings. And according to the question, the student observed that the temperature of the water increased during the course of the minutes.