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3 years ago

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How many milliliters of calcium with a density of 1.55 g/mL are needed to produce 12.4 g of hydrogen gas in the single replaceme

nt reaction below be sure to show the work that you did to solve the problem and balanced equation Ca + HF -> CaF2 + H2

1 answer:

timurjin [86]3 years ago

3 0

Answer:

Help

Explanation:

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Which does not show earths past environment?<br><br> Please hurry I need this :((

ANTONII [103]

There’s no options!!

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3 years ago

What is the equation for calculating gravitational potential energy on the Earth

DedPeter [7]

Gravitational potential energy = mass x g x height

(The g is measured in N/kg and is usually 9.8 or 10)

Full info in picture below:

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3 years ago

The dissociation of calcium carbonate has an equilibrium constant of Kp= 1.20 at 800°C. CaCO3(s) ⇋ CaO(s) + CO2(g)

Fed [463]

Explanation:

(a) Formula that shows relation between $K_{c}$ and $K_{p}$ is as follows.

$K_c = K_p \times (RT)^{-\Delta n}$

Here, $\Delta n$ = 1

Putting the given values into the above formula as follows.

$K_c = K_p \times (RT)^{-\Delta n}$

= $1.20 \times (RT)^{-1}$

= $\frac{1.20}{0.0820 \times 1073}$

= 0.01316

(b) As the given reaction equation is as follows.

$CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)$

As there is only one gas so ,

$p[CO_{2}] = K_{p}$ = 1.20

Therefore, pressure of $CO_{2}$ in the container is 1.20.

(c) Now, expression for $K_{c}$ for the given reaction equation is as follows.

$K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}$

= $\frac{x \times x}{(a - x)}$

= \frac{x^{2}}{(a - x)}[/tex]

where, a = initial conc. of $CaCO_{3}$

= $\frac{22.5}{100} \times 9.56$

= 0.023 M

0.0131 = $\frac{x^{2}}{0.023 - x}$

x = 0.017

Therefore, calculate the percentage of calcium carbonate remained as follows.

% of $CaCO_{3}$ remained = $(\frac{0.017}{0.023}) \times 100$

= 75.46%

Thus, the percentage of calcium carbonate remained is 75.46%.

3 0

4 years ago

37.2 liters of a gas has a pressure of 362.43 kPa at 46.5 °C. If the pressure increases to

Nataliya [291]

Answer:

Option A. 25.7 L.

Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial volume (V1) = 37.2 L

Initial pressure (P1) = 362.43 kPa

Initial temperature (T1) = 46.5 °C

Final pressure (P2) = 693.9 kPa

Final temperature (T2) = 149.2 °C

Final volume (V2) =.?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This can be achieved as shown below:

Temperature (K) = Temperature (°C) + 273

T (K) = T (°C) + 273

Initial temperature (T1) = 46.5 °C

Initial temperature (T1) = 46.5 °C + 273 = 319.5 K.

Final temperature (T2) = 149.2 °C

Final temperature (T2) = 149.2 °C + 273 = 422.2 K.

Step 3:

Determination of the new volume.

This can be obtained by using the general gas equation as shown below:

Initial volume (V1) = 37.2 L

Initial pressure (P1) = 362.43 kPa

Initial temperature (T1) = 319.5 K

Final pressure (P2) = 693.9 kPa

Final temperature (T2) = 422.2 K

Final volume (V2) =.?

P1V1/T1 = P2V2/T2

362.43×37.2/319.5 = 693.9 × V2/422.2

Cross multiply

319.5×693.9×V2 = 362.43×37.2×422.2

Divide both side by 319.5 × 693.9

V2 = (362.43×37.2×422.2)/(319.5×693.9)

V2 = 25.7 L

Therefore, the new volume is 25.7 L.

8 0

4 years ago

How much heat is required to change 100 g of ice (H2O) at 253 K to vapor (steam) at 393 K?

soldier1979 [14.2K]

<u>Answer:</u> The heat required will be 58.604 kJ.

<u>Explanation: </u>

To calculate the amount of heat required, we use the formula:

$Q= m\times c\times \Delta T$

Q= heat gained or absorbed = ? J

m = mass of the substance = 100 g

c = heat capacity of water = 4.186 J/g ° C

Putting values in above equation, we get:

$\Delta T={\text{Change in temperature}}=(393-253)K=140K=140^oC$

$Q=100g\times 4.186J/g^oC\times 140^oC$

Q = 58604 Joules = 58.604 kJ (Conversion factor: 1 kJ = 1000J)

Thus, heat released by 100 grams of ice is 58.604kJ.

3 0

4 years ago