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3 years ago

PLEASE HELP Is this a redox reaction? Zn + 2 HCl → ZnCl2 + H2

Chemistry

2 answers:

s2008m [1.1K]3 years ago

5 0

The answer will be number 4 if not i’m sorry

Alla [95]3 years ago

3 0

Answer:

answer is #4

Explanation:

Zn^0 -------> Zn^2+ ,

H^1+ --------> H^0

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Given 4.14 times 10^23 atoms of sodium, how many miles of sodium would you have

Cerrena [4.2K]

0.69moles

Explanation:

Given parameters:

Number of atoms = 4.14 x 10²³atoms

Atom = Na

Unknown:

Number of moles of Na atom in it = ?

Solution:

A mole of a substance is the quantity of substance that contains the avogadro's number of particles.

1 mole = 6.02 x 10²³particles

To solve this problem, simply use the expression below;

Number of moles = $\frac{number of particles }{6.02 x 10^{23} }$

Input the parameters;

Number of moles = $\frac{4.14 x 10^{23} }{6.02 x 10^{23} }$ = 0.69moles

learn more;

Number of moles brainly.com/question/1841136

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4 years ago

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FrozenT [24]

Answer: Not me

Explanation:

Not me

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3 years ago

Q2. Which of the following molecules may show.

jonny [76]

Answer:

I think is answer

CH3cl3

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3 years ago

The particles of a solid _____ in place.

chubhunter [2.5K]

Holds the particles tightly

7 0

3 years ago

Read 2 more answers

The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for

Zigmanuir [339]

Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

$\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]$

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

[A₀] = initial concentration

[A]= final concentration

t= time

k= rate constant

Half life: Half life is time to reduce the concentration of reactant of its half.

$t_{\frac{1}{2} }=\frac{0.693}{k}$

Here $t_{\frac{1}{2} }=0.13 min$

$k=\frac{0.693}{t_{\frac{1}{2}} }$

$\Rightarrow k=\frac{0.693}{13 }$

To find the time takes for it to decrease to 0.085 we use the below equation

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

$\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}$

Here , $k=\frac{0.693}{13 }$ , [A₀] = 0.13 m and [ A] = 0.085 M

$t=\frac{2.303}{\frac{0.693}{13} } log(\frac{0.13}{0.085})$

$\Rightarrow t= 7.97\approx 8.0$

Therefore it takes 8.0 mins for it to decrease to 0.085 M

7 0

3 years ago