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3 years ago

14

Select all that apply. Which equations are correct?

2 answers:

Elena-2011 [213]3 years ago

7 0

Sorry i can’t see it :(

antoniya [11.8K]3 years ago

3 0

Answer: Dang I wish I wasn't on a school chrome, that way I can see the photo without it getting blocked *cries*

Explanation:

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Zn + 2HCl -----> ZnCl2 + H2 <br> what mass of ZnCl2 will be produced if 25g of Hcl react

labwork [276]

What? It would be nice if you show pic

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3 years ago

Which of the following reasons (specific to this lab) would have caused the percent yield of calcium carbonate to be less than 1

mote1985 [20]

Answer:

Some of the calcium carbonate remained stuck to the side of the glassware after filtering

Some of the calcium carbonate spilled when measuring its mass

Explanation:

The percent yield is obtained from;

Actual yield/Theoretical yield * 100

Usually, the actual yield is less than the theoretical yield thereby making the percent yield less than 100%.

In the case of this particular reaction; when some calcium carbonate remains stuck on the glassware or some calcium carbonate is spilled during weighing, the actual yield of product obtained from the reaction becomes less than the theoretical or calculated yield hence the percent yield becomes less than 100%.

8 0

3 years ago

How many fluorine atoms bond with calcium to form calcium fluoride?

allsm [11]

2 atoms will bond with calcium to form Calcium Fluoride

3 0

3 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

What is the mass percent of oxygen in Nickel (III) sulfate?

NeX [460]

The mass percent of oxygen in Nickel( $III$ )sulfate is $23.68$ percent.

At first, we look up the atomic masses for the element from the periodic table.

The atomic masses are found to be:

Ni is $58.7$
S is $32$
O is $16$

Next, we determine how many grams of each element are present in one mole of $Ni_{2}(SO_{4})_{3}$

$117.4g(2 mol*58.7)$ of Ni
$96g(3 mol*32)$ of S
$192g(12 mol * 16.00)$ of O

The mass of one mole $Ni_{2}(SO_{4})_{3}$ is:

$117.4g+96g+192g=405.4g$

So, The mass percent of oxygen $=(96g/405.4)*100=0.2368*100%\\$

$=23.68 %$ percent.

To learn more about Mass percent, refer to:

brainly.com/question/25255030

3 0

2 years ago