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krok68 [10]
3 years ago
11

How much heat is released when 1.4 mol of hydrogen fluoride are produced?

Chemistry
1 answer:
Assoli18 [71]3 years ago
7 0

Answer:

+375.2 KJ.

Explanation:

The balanced equation for the reaction is given below:

H₂ + F₂ —> 2HF ΔH = +536 KJ

From the balanced equation above,

2 moles of HF required +536 KJ .

Finally, we shall determine the heat required to produce 1.4 mol of hydrogen fluoride, HF. This is illustrated below:

From the balanced equation above,

2 moles of HF required +536 KJ .

Therefore, 1.4 moles of HF will require = (1.4 × 536)/2 = +375.2 KJ

Thus, +375.2 KJ of heat energy is required.

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MakcuM [25]

Answer:

4 * 10^{-5}cm = 0.00004cm

Explanation:

Given

4 * 10^{-5}cm = [\ ]cm

Required

Solve

Apply law of indices

4 * \frac{1}{10^5}cm = [\ ]cm

Evaluate 10^5

4 * \frac{1}{100000}cm = [\ ]cm

4 * 0.00001cm = [\ ]cm

0.00004cm = [\ ]cm

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4 * 10^{-5}cm = 0.00004cm

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A cylindrical piece of metal is 4.5 dm in height with radius of 5.50 x 10^-5 km.
adell [148]

Answer:

a) V=4.3x10^3mL

b) V=4.3x10^6mm^3

c) \rho=1.5x10^5g/L

Explanation:

Hello,

a) In this case, the given height in cm is:

h=4.5dm*\frac{1m}{10dm}* \frac{100cm}{1m}=45cm

And the radius in cm is:

r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{100cm}{1m}=5.5cm

Thus, the volume in cubic centimeters which is also equal in mL (1cm³=mL) is:

V=\pi (5.5cm)^2*45cm\\\\V=4.3x10^3cm^3=4.3x10^3mL

b) In this case, the given height in mm is:

h=4.5dm*\frac{1m}{10dm}* \frac{1000mm}{1m}=450mm

And the radius in mm is:

r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{1000mm}{1m}=55mm

Thus, the volume in cubic millimeters is:

V=\pi (55mm)^2*450mm\\\\V=4.3x10^6mm^3

c) Finally, since 1000 mL equal 1 L, the required density in g/L turns out:

\rho=\frac{m}{V}=\frac{6.54x10^5g}{4.3x10^3mL}*\frac{1000mL}{1L}\\   \\\rho=1.5x10^5g/L

Best regards.

8 0
4 years ago
A gas sample occupies 200 mL at 760 mm Hg. What volume does the gas occupy at 400 mm Hg?
astra-53 [7]

Answer:

201.9

Explanation:

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