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denis-greek [22]
2 years ago
5

2. What is the final temperature when a 32.0 g piece of diamond at 33.5°C is heated with 360 J of energy?

Chemistry
1 answer:
Brums [2.3K]2 years ago
6 0

Answer:

55.6 °C

Explanation:

From the question given above, the following data were obtained:

Mass (M) of diamond = 32.0 g

Initial temperature (T₁) = 33.5°C

Heat (Q) required = 360 J

Specific heat capacity (C) of diamond = 0.509 J/gºC

Final temperature (T₂) =?

Next, we shall determine the change in temperature. This can be obtained as follow:

The final temperature can be obtained as follow:

Mass (M) of diamond = 32.0 g

Heat (Q) required = 360 J

Specific heat capacity (C) of diamond = 0.509 J/gºC

Change in temperature (ΔT ) =?

Q = MCΔT

360 = 32 × 0.509 × ΔT

360 = 16.288 × ΔT

Divide both side by 16.288

ΔT = 360 / 16.288

ΔT = 22.1 °C

Finally, we shall determine the final temperature. This can be obtained as follow:

Initial temperature (T₁) = 33.5°C

Change in temperature (ΔT ) = 22.1 °C

Final temperature (T₂) =?

ΔT = T₂ – T₁

22.1 = T₂ – 33.5

Collect like terms

22.1 + 33.5 = T₂

T₂ = 55.6 °C

Therefore, the final temperature is 55.6 °C.

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How many grams of NaCl (molecular weight = 58.4 g mole-1) would you dissolve in water to make a total volume of 500 mL of soluti
kolezko [41]

Given :

Volume , V = 500 mL .

Molarity , M = 0.5 M .

Molecular mass of NaCl is 58.4\ g/mole .

To Find :

How many grams of NaCl is required .

Solution :

Let , NaCl required is x gram .

Molarity is given by :

M=\dfrac{\text{Number of moles}}{\text{Volume (in liters) }}\\\\M=\dfrac{m}{M\times V}\\\\0.5=\dfrac{x}{58.4\times 0.5}\\\\x=0.5^2\times 58.4\ g\\\\x=14.6\ g

Hence , this is the required solution.

6 0
3 years ago
When methanol, CH 3 OH , is burned in the presence of oxygen gas, O 2 , a large amount of heat energy is released. For this reas
luda_lava [24]

<u>Answer:</u> The mass of methanol that must be burned is 24.34 grams

<u>Explanation:</u>

We are given:

Amount of heat produced = 581 kJ

For the given chemical equation:

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H=-764kJ

By Stoichiometry of the reaction:

When 764 kJ of heat is produced, the amount of methanol reacted is 1 mole

So, when 581 kJ of heat will be produced, the amount of methanol reacted will be = \frac{1}{764}\times 581=0.7605mol

To calculate mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of methanol = 0.7605 moles

Molar mass of methanol = 32 g/mol

Putting values in above equation, we get:

0.7605mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.7605mol\times 32g/mol)=24.34g

Hence, the mass of methanol that must be burned is 24.34 grams

4 0
3 years ago
Hydrogen is manufactured on an industrial scale by this sequence of reactions: Write an equation that gives the overall equilibr
RideAnS [48]

The question is incomplete. The complete question is :

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

The net reaction is  :

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 and K_2. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.

Solution :

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$     ...............(1)

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$  ...................(2)

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$

On multiplication of equation (1) and (2), we get

$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$

$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$  .................(4)

Comparing equation (3) and equation (4), we get

$K=K_1K_2$

4 0
2 years ago
I need help with this question
dsp73

Answer:

B?

Explanation:

it just makes the most sense in my head

7 0
3 years ago
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A weather balloon is inflated to a volume of 27.9L at a pressure of 732mmHg and a temperature of 30.1?C. The balloon rises in th
masya89 [10]

Answer:

45.4 L

Explanation:

Using Ideal gas equation for same mole of gas as

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

Given ,  

V₁ = 27.9 L

V₂ = ?

P₁ = 732 mmHg

P₂ = 385 mmHg

T₁ = 30.1 ºC

T₂ = -13.6  ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (30.1 + 273.15) K = 303.25 K  

T₂ = (-13.6 + 273.15) K = 259.55 K  

Using above equation as:

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

\frac{{732}\times {27.9}}{303.25}=\frac{{385}\times {V_2}}{259.55}

\frac{385V_2}{259.55}=\frac{20422.8}{303.25}

148225V_2=6729708.26018

Solving for V₂ , we get:

<u>V₂ = 45.4 L</u>

7 0
3 years ago
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