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3 years ago

Number 14? Please :)

Chemistry

1 answer:

ASHA 777 [7]3 years ago

4 0

Evaporated is the answer

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Molarity of 2%W/V of NaOH is

TiliK225 [7]

Assume it is 1 litre and weighs 1kg.
2 percent of 1 kg is 20g.
20g divided by molar mass of NaOH.
20g divide by 40 = 0.5 mole
0.5 mole in a litre would be 0.5M
That is the answer: 0.5M

6 0

3 years ago

Which is one characteristic that temperate marine and temperate continental climates have in common?

shepuryov [24]

Answer:

it doesn't have savanna climate which is mostly found in Africa content

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3 years ago

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Flowers start as a sprout. Opinion Fact

Harman [31]

Answer:

FACT!!!!!!!!!!

Explanation:

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3 years ago

Which statement best describes balancing equations and the law of conservation of mass?

Marta_Voda [28]

Answer:

B.) The number of atoms is the same in the reactants and in the products, and the total mass is the same in the reactants and in the products.

Explanation:

law of conservation of matter/mass states that no atoms can be created or destroyed

hope this helps :)

6 0

3 years ago

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl ( aq ) HCl(aq) , as

Daniel [21]

Answer:

1.368 grams of manganese dioxide must b added to HCl to obtain 385 mL of chlorine gas.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of chlorine gas = $765 Torr=\frac{765}{760}atm$

1 atm = 760 Torr

V = Volume of chlorine gas = 385 mL = 0.385 L ( 1 mL - 0.001 L)

n = number of moles of chlorine gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of chlorine gas =25 °C= 25 + 273 K = 300 K

Putting values in above equation, we get:

$(\frac{765 }{760}atm)\times 0.385 L=n\times (0.0821L.atm/mol.K)\times 300K\\\\n=0.01573 mole$

$MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)$

According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese dioxide ,then 0.01573 moles of chlorine gas will be obtained from :

$\frac{1}{1}\times 0.01573 mol=0.01573 mol$ of manganese dioxide

Mass of 0.01573 moles of manganese dioxide:

0.01573 mol × 86.94 g/mol = 1.368 g

1.368 grams of manganese dioxide must b added to HCl to obtain 385 mL of chlorine gas.

5 0

4 years ago