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EleoNora [17]
3 years ago
7

A chemist performs the following reaction to make sodium nitrate

Chemistry
2 answers:
vagabundo [1.1K]3 years ago
5 0

Answer:

See below

Explanation:

A chemist preforms the following: Neutralizing nitric acid (HNO3) with soda ash (Na2CO3). This reaction will then create sodium nitrate and carbonic acid, which will then decompose into water (H20)

ladessa [460]3 years ago
4 0

Answer:

The salt (Sodium Nitrate) is prepared by the reaction of nitric acid and sodium hydroxide. The reaction is also known as neutralization reaction.

Explanation:

The reaction between nitric acid (HNO_{3}) and sodium hydroxide (NaOH) will result in the formation of salt (Sodium nitrate). The balanced reaction between nitric acid and sodium hydroxide is shown below

\textrm{HNO}_{3}\left ( aq \right )+\textrm{NaOH}\left ( aq \right )\rightarrow \textrm{NaNO}_{3}+\textrm{H}_{2}\textrm{O}\left ( l \right )

HNO_{3} is a strong acid and NaOH is a strong base. This reaction is an example of neutralization reaction.

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An unknown gas Q requires 2.67 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the
Misha Larkins [42]

Answer:

The correct answer is 199.66 grams per mole.

Explanation:

Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,  

R1/R2 = √ M2/√ M1

Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.  

Rate Q/Rate N2 = √M of N2/ √M of Q

The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67

Now putting the values we get,  

rate of N2/2.67/rate of N2 = √28/ √M of Q

√M of Q = √ 28 × 2.67

M of Q = (√ 28 × 2.67)²

M of Q = 199.66 grams per mole

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