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Sliva [168]
2 years ago
15

A______Occouring________Solid that has a crystal structure and a definite______composition.

Chemistry
1 answer:
postnew [5]2 years ago
7 0

Answer:

What are the answers.

Explanation:

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The balanced chemical equation for the reaction between sodium chloride and silver nitrate is:
bonufazy [111]

Answer:

We can Interprete it as 1mole of Sodium Chloride and 1mole of Silver Nitrate React to Produce

1Mole of Silver Chloride and 1Mole of Sodium Nitrate

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A vehicle is moving with a velocity of 20 m/s.how far does it moves in one hrs
Airida [17]
Distance= Speed x Time

1hr=60mins

60mins=3600 secs

20 x 3600=72000

72000 is your answer.

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3 years ago
Crushing large chunks of sodium chloride:<br> What lab equipment
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A first-order reaction has a half-life of 16.7 s . How long does it take for the concentration of the reactant in the reaction t
andrew11 [14]

It takes 33.4 s for the concentration of A to fall to one-fourth of its original value.

A <em>half-life</em> is the time it takes for the concentration to fall to half its original value.

Assume the initial concentration is 1.00 mol/L. Then,

\text{1.00 mol/L }\stackrel{\text{1st half-life }}{\longrightarrow}\text{ 0.50 mol/L } \stackrel{\text{ 2nd half-life }}{\longrightarrow}\text{0.25 mol/L}\\

The concentration drops to one-fourth of its initial value in two half-lives.

∴ Time = 2 × 16.7 s = 33.4 s

5 0
2 years ago
The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat
Valentin [98]

Answer:

(a)

0.0342M

(b)

t_{1/2}=17.36s\\t_{1/2}=23.15s

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\

[NOBr]=\frac{1}{29.2/M}=0.0342M

(b) Now, for a second-order reaction, the half-life is computed as shown below:

t_{1/2}=\frac{1}{k[NOBr]_0}

Therefore, for the given initial concentrations one obtains:

t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s

Best regards.

8 0
2 years ago
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