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3 years ago

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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At a certain concentration of H2 and I2, the initial rate of reaction is 4.0 x 104 M / s. What would the initial rate of the rea

andreev551 [17]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

$rate=k[H_2]^2[I_2]^2$

At a certain concentration of $H_2$ and $I_2$ , the initial rate of reaction is 4.0 × 10⁴ M/s. What would the initial rate of the reaction be if the concentration of

Answer : The initial rate of the reaction will be, $1.0\times 10^4M/s$

Explanation :

Rate law expression for the reaction:

$rate=k[H_2]^2[I_2]^2$

As we are given that:

Initial rate = 4.0 × 10⁴ M/s

Expression for rate law for first observation:

$4.0\times 10^4=k[H_2]^2[I_2]^2$ ....(1)

Expression for rate law for second observation:

$R=k(\frac{[H_2]}{2})^2[I_2]^2$ ....(2)

Dividing 2 by 1, we get:

$\frac{R}{4.0\times 10^4}=\frac{k(\frac{[H_2]}{2})^2[I_2]^2}{k[H_2]^2[I_2]^2}$

$\frac{R}{4.0\times 10^4}=\frac{1}{4}$

$R=1.0\times 10^4M/s$

Therefore, the initial rate of the reaction will be, $1.0\times 10^4M/s$

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3 years ago

Which Step In the nitrogen cycle is accelerated at the beginning of the eutrophication process

krek1111 [17]

The step in the nitrogen that is accelerated at the beginning of the eutrophication process is producer uptake of nitrogen. The correct answer is C.

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3 years ago

Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constan

ivolga24 [154]

Answer : The solubility is, $1.4\times 10^{-5}mol/L$

Explanation : Given,

$k_H$ = Henry's law constant of argon = $1.5\times 10^{-3}mol/L.atm$

First we have to calculate the pressure of argon.

Pressure of argon = $1.0atm\times \frac{0.93}{100}$

Pressure of argon = 0.0093 atm

Now we have to calculate the solubility.

As, the solubility of argon in 1 atm pressure = $1.5\times 10^{-3}mol/L$

So, the solubility of argon in 0.0093 atm pressure = $\frac{0.0093atm}{1atm}\times 1.5\times 10^{-3}mol/L$

= $1.4\times 10^{-5}mol/L$

Thus, the solubility is, $1.4\times 10^{-5}mol/L$

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3 years ago

Jet001 [13]

Answer:

1 mol SO2 contains 6.0213*10^23 molecules

6.023*10^24 molecules = 10 mol SO2

Equation

S(s) + O2(g) → SO2(g)

1 mol S reacts with 1 mol O2 to prepare 1 mol SO2

To prepare 10 mol SO2 you require : 10 mol S plus 10 mol O2

And that is the answer to the question

If you want a mass :

Molar mass S = 32 g/mol You require 10 mol = 320 g

Molar mass O2 = 32 g/mol :You require 10 mol = 320 g

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3 years ago

Read 2 more answers

List two products derived from ethylene

laiz [17]

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2 years ago