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Airida [17]
3 years ago
7

What would be the MOLARITY (M) of the NaBr in the same buffer solution, if its concentration is now 0.9% (w/v)?

Chemistry
1 answer:
IRINA_888 [86]3 years ago
6 0
If by buffer solution you mean, a solution of the same concentration, the molarity (M) would stay the same. You would just have a greater volume.
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Nitric acid is often manufactured from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming of natural
hoa [83]

<u>Answer:</u> The net chemical equation is given below.

<u>Explanation:</u>

To determine the net chemical equation, we will look into the intermediate steps:

<u>Step 1:</u> Formation of ammonia from hydrogen and nitrogen gas.

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)        ......(1)

<u>Step 2:</u> Formation of nitric acid from ammonia and oxygen gas.

NH_3(g)+2O_2(g)\rightarrow HNO_3(g)+H_2O(g)      .....(2)

The net chemical equation for the formation of nitric acid from hydrogen gas, nitrogen gas and oxygen gas follows by adding both the equations, we get:

N_2(g)+2O_2(g)+3H_2(g)\rightarrow HNO_3(g)+H_2O(g)+NH_3(g)

By Stoichiometry of the reaction:

1 mole of hydrogen gas reacts with 2 moles of oxygen gas and 3 moles of hydrogen gas to produce 1 mole of nitric acid, 1 mole of water and 1 mole of ammonia.

Hence, the net chemical equation is given below.

4 0
3 years ago
A first-order reaction has a half-life of 20.0 minutes. Starting with 1.00 × 1020 molecules of reactant at time t = 0, how many
zalisa [80]

Half-life time of a reaction is time at which reactant concentration becomes half of its initial value.

Half-life of the first order reaction is 20 min. Rate constant can be calculated as follows:

K=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{20 min}=0.03466 min^{-1}

The rate expression for first order reaction is as follows:

k=\frac{2.303}{t}log\frac{A_{0}}{A_{t}}

initial number of molecules of reactant are 10^{20}, time is 100 min thus, putting the values to calculate number of reactant at time 100 min,

0.03466 min^{-1}=\frac{2.303}{100 min}log\frac{[10^{20}]}{A_{t}}

On rearranging,

\frac{10^{20}}{A_{t}}=31.988

Or,

A_{t}=3.13\times 10^{18}

Therefore, number of molecules unreacted will be 3.13\times 10^{18}

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Answer:303.15

Explanation:

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I’m confused on this concept please help asap!
mixas84 [53]
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Answer:

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