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vivado [14]
3 years ago
5

Pls help me I beg you

Chemistry
2 answers:
just olya [345]3 years ago
6 0

Answer:

Es la c

Explanation:

Espero que te ayude

nexus9112 [7]3 years ago
5 0

Answer:

D. Nathan applies enough force through pushes to move the piano

Explanation:

What moves the piano is a net force that is applied through the effort of Nathan on the instrument.

Force is a pull or push on a body.

When force is applied on a body, it causes the state of motion of the body to change.

Therefore, if a body has a net force greater than zero, the body will in likewise begin to move. As Nathan pushes on the piano, it will eventually move.

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Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4
inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

M=\frac{n_{solute}}{V_{solution}}

So you can solve for the moles of the solute:

n_{solute}=M*V_{solution}

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

n_{solute}=1mol/L*1.5L=1.5mol

But this is just a supposition.

Regards.

4 0
2 years ago
Adenosine triphosphate is the:
Nadya [2.5K]
C. Cellular respiration
7 0
3 years ago
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The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s
raketka [301]

Answer : The equilibrium concentration of H_3O^+ in the solution is, 2.1\times 10^{-3}M

Explanation :

The dissociation of acid reaction is:

                       C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-

Initial conc.        c                                 0                0

At eqm.             c-x                                 x                x

Given:

c = 7.0\times 10^{-2}M

K_a=6.3\times 10^{-5}

The expression of dissociation constant of acid is:

K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}

K_a=\frac{(x)\times (x)}{(c-x)}

Now put all the given values in this expression, we get:

6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}

x=2.1\times 10^{-3}M

Thus, the equilibrium concentration of H_3O^+ in the solution is, 2.1\times 10^{-3}M

4 0
3 years ago
1/4 checkpoint
DochEvi [55]

Answer:

here's the answers

Explanation:

7 0
2 years ago
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A 0.227 mol chunk of dry ice (solid CO2) changes to gas. What is the volume of that gas measured at 27 °C and 740 mmHg?
jeka57 [31]

Answer:

Explanation:3.2 ft 2 fti2 ft 4 ft ft2

4 0
2 years ago
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