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3 years ago

What happens when potassium metal react with nitric acid

Chemistry

1 answer:

Marrrta [24]3 years ago

3 0

Answer:

Astamañana vaya buenas noches

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Integrated rate law for second order unimolecular irreversible

kirill115 [55]

Answer:

The rate law for second order unimolecular irreversible reaction is

$\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }$

Explanation:

A second order unimolecular irreversible reaction is

2A → B

Thus the rate of the reaction is

$v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}$

rearranging the ecuation

$-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}$

Integrating between times 0 to <em>t </em>and between the concentrations of $[A]_{0}$ to <em>[A].</em>

$\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}$

Solving the integral

$\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }$

5 0

3 years ago

A solid precipitate is formed from the reaction of lead nitrate and sodium chromate.

Serhud [2]

I believe it's Filtration.

4 0

3 years ago

Read 2 more answers

Nal+ pb(SO4) 2 Pbl4 + Na2SO4 explain how to balance the chemical equation and classify its reaction type

Pani-rosa [81]

Explanation:

because there are 4 Iodines on the left, we'll put. 4 in front of NaI to balance it. This would result in 4 Na on the left, so we'll put a 2 in front of Sodium Sulfate to balance the right side. Now we have 4 Na and I on both side, as well as 2 Sulfate on both sides. Pb is already balanced. The equation is now complete.

3 0

3 years ago

Calculate the concentration of acetate ion in a buffer solution made from 2.00 mL of 0.50 M acetic acid and 8.00 mL of 0.50 sodi

Lelu [443]

Answer:

1 M

Explanation:

Equation of reaction is;

CH3COOH + CH3COONa -------------------> 2CH3COO^- + NaH

1 moles of each of the reactants react to give 2 moles of the acetate ion.

From the question, we have that 2.00 mL that is (2÷1000)L of 0.50 M acetic acid reacted with 8.00 mL that is (8/1000)L of 0.50 sodium acetate.

Then from equation, n = CV -------------------------------------------(1).

Where n= number of moles, V= volume, C= concentration.

Number of moles,n of acetic acid = 0.50M× 2/1000L.

n(acetic acid)= 0.001 moles.

Number of moles,n of sodium acetate= 0.50M ×(8/1000)L.

n(sodium acetate)= 0.004 moles.

0.001 moles of acetic acid react with 0.004 moles of Sodium acetate

Therefore, acetic acid is the limiting reagent.

One mole of acetic acid produces 2 moles of acetate ion.

0.001 mole of acetic acid produces= 0.002 moles of acetate ion.

Using the equation (1) that is, n= CV.

0.002= C× 2/1000

C= 0.002/0.002

C= 1 M

8 0

4 years ago

How many moles of H2 and N2 can be formed by the decomposition of 0.145 mol of ammonia, NH3 ?

zepelin [54]

Answer:

Explanation:

The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.

5 0

3 years ago