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4 years ago

15

Identify Cobalt sulfride

1 answer:

jeyben [28]4 years ago

7 0

It has cobalt and sulfur

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29. Which alcohol combines with carboxylic acid to produce the ester called ethyl butanoate?

olya-2409 [2.1K]

Answer:

The answer is option D.

ethanol

Hope this helps you

8 0

3 years ago

For two variables in a direct proportion, what is the result of doubling one variable? the other variable is doubled the other v

Aliun [14]

Other variable is doubled.

8 0

3 years ago

Read 2 more answers

a sample of cobalt (specific heat of Co=0.418J/G C )at 100.0 C is dropped into a calorimeter containing 500.0 mL of water at 21.

Alex17521 [72]

Answer:

$m_{Co}=6998g=7.0kg$

Explanation:

Hello there!

In this case, according to this equilibrium temperature problem, we can set up the following equation to relate the mass, specific heat and temperature change:

$Q_{Co}=-Q_{w}\\\\m_{Co}C_{Co}(T_f-T_{Co})=-m_{w}C_{w}(T_f-T_{w})$

Thus, we solve for the mass of cobalt as shown below:

$m_{Co}=\frac{-m_{w}C_{w}(T_f-T_{w})}{C_{Co}(T_f-T_{Co})} \\\\m_{Co}=\frac{-500.00g*4.184J/g\°C(67.1\°C-21.1\°C)}{0.418J/g\°C(67.1\°C-100\°C)} \\\\m_{Co}=6998g=7.0kg$

Best regards!

7 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

ludmilkaskok [199]

Answer: It would be C

7 0

3 years ago