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STatiana [176]
3 years ago
11

A gas at 1.2 atm has a

Chemistry
1 answer:
Papessa [141]3 years ago
8 0

Answer:

1.26 mole

Explanation:

Using ideal gas law

PV=nRT

R= 0.08206

T=273+87=360

1.2×31=n×0.08206×360

37.2 =n × 29.54

n= 1.26 mole

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Answer:

D. 2NaBr + Cl_2\rightarrow 2NaCl + Br_2

Explanation:

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In this case, for the given set of chemical reactions, it is possible to infer that D. is a categorized as redox due to the following:

Since both chlorine and bromine remain as diatomic gases, their oxidation states in such a form is 0, but as anions with lithium cations they have a charge of - according to the following reaction and half-reactions:

2NaBr + Cl_2\rightarrow 2NaCl + Br_2

Cl_2^0+2e^-\rightarrow 2Cl ^-\\\\2Br^- \rightarrow  Br_2^0+2e^-

Unlike the other reactions whereas no change in the oxidation states is evidenced.

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Answer:

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A flask contains 6g hydrogen gas and 64 g oxygen at rtp the partial pressure of hydrogen gas in the flask of the total pressure
Alex

Answer:

B.3/5p

Explanation:

For this question, we have to remember <u>"Dalton's Law of Partial Pressures"</u>. This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.

Additionally, we have a <em>proportional relationship between moles and pressure</em>. In other words, more moles indicate more pressure and vice-versa.

P_i=P_t_o_t_a_l*X_i

Where:

P_i=Partial pressure

P_t_o_t_a_l=Total pressure

X_i=mole fraction

With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:

<u>moles of hydrogen gas</u>

The molar mass of hydrogen gas (H_2) is 2 g/mol, so:

6g~H_2\frac{1~mol~H_2}{2~g~H_2}=~3~mol~H_2

<u>moles of oxygen gas</u>

The molar mass of oxygen gas (O_2) is 32 g/mol, so:

64g~H_2\frac{1~mol~H_2}{32~g~H_2}=~2~mol~O_2

Now, total moles are:

Total moles = 2 + 3 = 5

With this value, we can write the partial pressure expression for each gas:

P_H_2=\frac{3}{5}*P_t_o_t_a_l

P_O_2=\frac{2}{5}*P_t_o_t_a_l

So, the answer would be <u>3/5P</u>.

I hope it helps!

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