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My name is Ann [436]
3 years ago
14

If 6.35 g of cu(no3)2 is dissolved in water to make a 0.750 m solution, what is the volume of the solution (in ml)?

Chemistry
1 answer:
miv72 [106K]3 years ago
7 0
First, we need to find out how many moles we have of Cu ( NO_{3} )_{2} if we are given 6.35g.

We need to calculate the molar mass of the molecule:

Cu: 63.546g
N: 14.01g * 2 = 28.02g
O: 15.999g * 6 = 95.994

Add them together to get: 187.56g

So now we need to find the number of moles we are given. This can be done by setting up a multiplication equation and cancelling the units:

\frac{1mole}{187.56g}* \frac{6.35g}{1}=0.034 mol _{Cu( NO_{3} )_{2}   }

So now that we know the number of moles that we have, we can use the molarity equation to find the volume of the solution. The molarity equation is:

Molarity= \frac{moles}{Liters}

So since we are given the molarity in the equation (0.750 M), we can plug in the molarity and moles to find the volume of the solution:

0.750= \frac{0.034mol}{volume(L)}

volume(L)=0.04533L

So now we know that the volume is 0.04533 Liters; however the question asks for the answer in milliliters. To do this, we must multiply by 1000:

\frac{0.045L}{1} * \frac{1000mL}{1L}=45.33mL

So the volume of the solution (<em>in mL</em>) is 45.33 mL.
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How many moles of water, H2O contain 2.0x10^22 molecules of water
Licemer1 [7]
I mole of water has an Avogadro number of molecules.

1 mole = 6.02 * 10^ 23  molecules.

6.02 * 10^ 23  molecules  =  1 mole of water
 
1 molecule                        =  1/(6.02 * 10^23)  mole of water

2.0 * 10^22 molecules would have   =    (2*10^22) * 1/(6.02*10^23)

                                                             = 0.033

2* 10 ^22 molecules of water would have 0.033 moles of water.
5 0
3 years ago
A ball has a 17 j of kinetic energy and its mechanical energy is 25 j. Calculate its potential energy
ycow [4]

The potential energy of the ball is 8 J.

<h3>What is mechanical energy?</h3>

Mechanical energy is given as the energy possessed by an object to do work. It is given as the sum of the kinetic energy and potential energy.

Mechanical energy = Kinetic energy + Potential energy

The given ball has:

  • Mechanical energy = 25 J
  • Kinetic energy = 17 J

The potential energy is given as;

25 J = 17 J + Potential energy

Potential energy = 25 J - 17 J

Potential energy =  8 J

The potential energy of the ball is 8 J.

Learn more about potential energy, here:

brainly.com/question/24284560

#SPJ4

3 0
2 years ago
Hey guys, I was wondering if you could give me an explanation on aqueous electrolysis.
Crazy boy [7]
Aqueous compounds are a mixtures of two electrolytes – the compound and water.

Does this help a little?
6 0
3 years ago
One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c
Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of \rm O_2\; (g) is larger than that of \rm H_2\; (g) (by a factor of about 16.) Therefore, the mass of the \rm O_2\; (g) sample is significantly larger than that of the \rm H_2\; (g) sample.

Explanation:

The \rm O_2\; (g) and the \rm H_2\; (g) sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles (\rm O_2\; (g) and \rm H_2\; (g) molecules, respectively.) That is:

n(\mathrm{O_2}) = n(\mathrm{H}_2).

Note that the mass of a gas m is different from the number of gas particles n in it. In particular, if all particles in this gas have a molar mass of M, then:

m = n \cdot M.

In other words,

  • m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2}).
  • m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2}).

The ratio between the mass of the \rm O_2\; (g) and that of the \rm H_2\; (g) sample would be:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}.

Since n(\mathrm{O_2}) = n(\mathrm{H}_2) by Avogadro's Law:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}.

Look up relative atomic mass data on a modern periodic table:

  • \rm O: 15.999.
  • \rm H: 1.008.

Therefore:

  • M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}.
  • M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}.

Verify whether \begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}:

  • Left-hand side: \displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1.
  • Right-hand side: \displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9.

Note that the mass of the \rm H_2\; (g) sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

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4 years ago
Which of the following human activities can increase the risks of flooding?
adelina 88 [10]

deforestation i hope this helps

7 0
3 years ago
Read 2 more answers
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