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elena-14-01-66 [18.8K]
3 years ago
9

The density of lead is 11.3 g/cm. What is the mass of a cube of lead with 10.0 cm on each edge? Show work please.​

Chemistry
1 answer:
tresset_1 [31]3 years ago
4 0

Mass of the lead cube :

  • \boxed{ \boxed{11.3 \:  \: kg}}

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Assume that temperature and number of moles of gas are constant in this problem.
IrinaVladis [17]

Question:

a. a direct linear relationship

b. an inverse linear relationship

c. a direct nonlinear relationship

d. an inverse nonlinear relationship

Answer:

The correct option is;

d. An inverse nonlinear relationship

Explanation:

From the universal gas equation, we have;

P·V = n·R·T

Where we have the temperature, T and the number of moles, n constant, therefore, we have

P×V = Constant, because, R, the universal gas constant is also constant, hence;

P×V = C

P = \frac{C}{V}

Since P varies with V then the graphical relationship will be an inverse nonlinear as we have

V P    C

1 5         5

2 2.5      5

3 1.67     5

4 1.25     5

5 1          5

6 0.83    5

7 0.7      5

8 0.63    5

9 0.56    5

10 0.5      5

Where:

V = Volume

P = Pressure

C = Constant = 5

P = C/V

The graph is attached.

4 0
3 years ago
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
WILL GIVE BRAINLIEST NO FAKE ANSWERS PLEASE!
AleksAgata [21]

Answer:

A) (3.2g)

Explanation:

Did you reposed this? Because I remember answering this

3 0
3 years ago
Read 2 more answers
Find the density of an object that has a mass of 12.69 grams and a volume of 3.5cm3
lutik1710 [3]

Answer:

3.62 g/cm³

Explanation:

density = mass ÷ volume

Therefore, do 12.69 divided by 3.5

6 0
3 years ago
How many grams of o2 are required to produce 358.5 grams of zno? 2zn + o2 ® 2zno?
Murrr4er [49]
The balanced equation for the reaction is ;
2Zn + O2 —> 2ZnO
The stoichiometry of O2 to ZnO is 1:2
The mass of ZnO formed - 358.5 g
The number of moles formed - 358.5 g / 81.4 g/mol = 4.4 moles
Therefore number of O2 moles reacted = 4.4 moles /2 = 2.2 mol
Mass of O2 reacted = 2.2 mol x 32 g/mol = 70.4 g
6 0
3 years ago
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