Which statement best describes an effect of tension acting on Earthis crudo

Students working in lab accidentally spilled 17 l of 3.0 m h2so4 solution. they find a large container of acid neutralizer that

jek_recluse [69]

Answer is: 8568.71 of baking soda.

Balanced chemical reaction: H₂SO₄ + 2NaHCO₃ → Na₂SO₄ + 2CO₂ + 2H₂O.

V(H₂SO₄) = 17 L; volume of the sulfuric acid.

c(H₂SO₄) = 3.0 M, molarity of sulfuric acid.

n(H₂SO₄) = V(H₂SO₄) · c(H₂SO₄).

n(H₂SO₄) = 17 L · 3 mol/L.

n(H₂SO₄) = 51 mol; amount of sulfuric acid.

From balanced chemical reaction: n(H₂SO₄) : n(NaHCO₃) = 1 :2.

n(NaHCO₃) = 2 · 51 mol.

n(NaHCO₃) = 102 mol, amount of baking soda.

m(NaHCO₃) = n(NaHCO₃) · M(NaHCO₃).

m(NaHCO₃) = 102 mol · 84.007 g/mol.

m(NaHCO₃) = 8568.714 g; mass of baking soda.

8 0

3 years ago

describe the difference between a monoprotic acid and a triprotic acid. How does a triprotic acid generate hydrogen ions?

natita [175]

With the Bronsted-Lowry theory of acids and bases, acids donate protons to other substances. A monoprotic acid is an an acid that can only supply one proton (hydrogen). A triprotic acid is an an acid that can supply up to three protons (hydrogens). A triprotic acid get get its hydrogen ions by receiving them from a base.

6 0

4 years ago

Imagine disease kills 85 percent of the wolf population. How will it affect the other organisms?

stealth61 [152]

Answer:

There would most likely be an overpopulation of the animals the wolfs ate.

Explanation:

7 0

3 years ago

Read 2 more answers

How many atoms are in one molecule of Iron II sulfate

dezoksy [38]

Answer:
The number of atoms contained by one molecule of Iron (II) Sulfate are 6.

Explanation:
Iron (II) Sulfate is mage up of two parts. One is the Positive part which constitutes of Fe⁺² and a negative part which constitutes of a polyatomic anion i.e. SO₄²⁻. As there are four Oxygen and one sulfur atom in sulfate Ion so sulfate ion contains 5 atoms in total. Therefore, five atoms from sulfate iona dn one atom of Iron ion makes a total of 6 atoms in one molecule of Iron (II) Sulfate.

5 0

3 years ago

If the value of Kc for the reaction is 2.50, what are the equilibrium concentrations if the reaction mixture was initially 0.500

ratelena [41]

Answer: The equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

Explanation:

We are given:

Initial concentration of sulfur dioxide = 0.500 M

Initial concentration of nitrogen dioxide = 0.500 M

Initial concentration of sulfur trioxide = 0.00500 M

Initial concentration of nitrogen monoxide = 0.00500 M

The chemical reaction follows:

$SO_2+NO_2\rightleftharpoons SO_3+NO$

Initial: 0.500 0.500 0.005 0.005

At eqllm: 0.500-x 0.500-x 0.005+x 0.005+x

The expression of equilibrium constant for the above reaction follows:

$K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}$

We are given:

$K_c=2.50$

Putting values in above equation, we get:

$2.50=\frac{(0.005+x)\times (0.005+x)}{(0.500-x)\times (0.500-x)}\\\\x=0.304,1.37$

Neglecting the value of x = 1.37, because change cannot be greater than the initial concentration

So, equilibrium concentration of sulfur dioxide = $(0.500-x)=(0.500-0.304)=0.196M$

Equilibrium concentration of nitrogen dioxide = $(0.500-x)=(0.500-0.304)=0.196M$

Equilibrium concentration of sulfur trioxide = $(0.00500+x)=(0.00500+0.304)=0.309M$

Equilibrium concentration of nitrogen monoxide = $(0.00500+x)=(0.00500+0.304)=0.309M$

Hence, the equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

4 0

3 years ago