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Paraphin [41]
3 years ago
13

What is the mass of 2.5 moles of hydrogen fluoride gas HF

Chemistry
1 answer:
Nady [450]3 years ago
7 0

Answer:

You will get 5.0 g of hydrogen.

Explanation:

As with any stoichiometry problem, we start with the balanced equation.

Sn

l

+

2HF

→

SnF

2

+

H

2

Moles of H

2

=

2.5

mol Sn

×

1 mol H

2

1

mol Sn

=

2.5 mol H

2

Mass of H

2

=

2.5

mol H

2

×

2.016 g H

2

1

mol H

2

=

5.0 g H

2

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The total pressure of gases a, b, and c in a closed container is 4.1 atm. if the mixture is 36% a, 42% b, and 22% c by volume, w
Natali [406]

 The  partial  pressure of  gas C  is 0.902  atm


  calculation

partial pressure of gas c  =[( percent by volume of  gas   C /  total  percent)   x total pressure]


percent  by  volume of gas C= 22%

Total   percent  = 36% +42%  + 22%  = 100 %

Total  pressure  =  4.1 atm


partial  pressure  of gas C  is therefore =  22/100 x 4.1 atm = 0.902  atm

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4 years ago
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Which factors cause a rising tide? A. The moon's gravity and Earth's rotation B. Density differences due to salinity and tempera
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If the observed value for a density is 0.80 g/mL and the accepted value is 0.70 g/mL what is the percent error?
kvv77 [185]

Answer:

<h2>The answer is 14.29 %</h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

actual density = 0.70 g/mL

error = 0.8 - 0.7 = 0.1

So we have

P(\%) =  \frac{0.1}{0.7}  \times 100 \\  = 14.285714...

We have the final answer as

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4 years ago
Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a
GrogVix [38]

Answer:

The calorie content of  65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

 ⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium)  + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

⇒ with mass of the cup = 0.624 kg

⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41  * (65/15) = 326.78 cal

8 0
3 years ago
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