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Anna35 [415]
3 years ago
10

What is the volume of 8.50 moles of hydrogen gas?

Chemistry
1 answer:
Nostrana [21]3 years ago
5 0
Do dimensional Analysis

8.5mol   22.4L
__         *_             = 190.4 L
1             1 mol

It's as if you are multiplying 8.5 moles by 1 because in every gas there is 22.4 Liters per mole
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calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w
kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is -22Jmole^-^1k^-^1

(b) The normal boiling point of water (J·K−1·mol−1) is -109Jmole^-^1K^-^1

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      (\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p = -5_m(\beta )+5_m(\alpha ) =  -\frac{\Delta H}{T}

 \alpha ,\beta → phases

ΔH → enthalpy of transition

T → temperature transition

 (\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p == -\frac{\Delta_fH}{T_f}

            = \frac{-6.008kJ/mole}{273.15K} ( \Delta_fH is the enthalpy of fusion of water)

           = -22Jmole^-^1k^-^1

(b) (\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}

                                  = \frac{40.656kJ/mole}{373.15K} (\Delta_v_a_p_o_u_rH is the enthalpy of vaporization)

                               = -109Jmole^-^1K^-^1

(c) \Delta\mu =\Delta\mu(l)-\Delta\mu(s) =-S_m\DeltaT

[\mu(l-5°C)-\mu(l,0°C)] =  [\mu(s-5°C)-\mu(s,0°C)]=-S_mΔT

\mu(l,-5°C)-\mu(s,-5°C)=-Sm\DeltaT [\mu(l,0

\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)

     = 109J/mole

6 0
2 years ago
CH, +<br> ,<br> 0, → _CO, +<br> Н,0<br> Balance chemical equation
Flura [38]

Answer:

2 CH2 + 3 O2 = 2 CO2 + 2 H2O

Explanation:

This is what I think that you meant by the question listed. When balancing a chemical equation, you want to make sure that there are equal amounts of each element on each side.

Originally, the equation's elements looked like this: 1 C on left & 1 C on right; 2 H on left & 2 H on right; 2 O on left and 3 O on right. Because these are not balanced, you need to add coefficients.

When adding coefficients, you need to make sure that all of the elements stay balanced, not just one that you are trying to fix. I know that some equations are really difficult to balance, and when that is the case, there are equation balancing websites that can help out.

However, what always helps me is making a chart and continuing to keep up with the changes I am making. It is a trial and error process.

6 0
3 years ago
Pls help!<br> Is volume equal to mass over density?
olganol [36]

density = mass/volume

volume= mass/density

Yes, you're correct.

5 0
3 years ago
Read 2 more answers
Suppose you have a dozen carbon atoms, a dozen gold atoms, and a dozen iron atoms. Even though you have the same number of each,
mixer [17]

Answer:

By weight they have the same mass, but the number of atoms is different

Explanation:

3 0
3 years ago
Read 2 more answers
The radius of a vanadium atom is 130 pm. How many vanadium atoms would have to be laid side by side to span a distance of 1.30 m
monitta

Answer:

5 000 000 (5 million atoms)

Explanation:

Let us assume that a vanadium atom has a spherical shape.

diameter of a sphere = 2 x radius of the sphere

Thus,

Radius of a vanadium atom = 130 pm

                                              = 130 x 10^{-12} m

The diameter of a vanadium atom = 2 x radius

                                                         = 2 x 130 x 10^{-12}

                                               = 260 x 10^{-12} m

Given a distance of 1.30 mm = 1.30 x 10^{-3} m,

The number of vanadium atoms required to span the distance = \frac{1.3*10^{-3} }{260*10^{-12} }

                                                  = 5000000

Therefore, the number of vanadium atom that would span a distance of 1.30 mm is 5 million.

3 0
3 years ago
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