True or false?<br><br> Freshwater can occur as a gas, liquid, or solid.

How many grams of sodium metal must be introduced to water to produce 3.3 grams of hydrogen gas?

Cloud [144]

Answer:

The mass of sodium metal that must be introduced to water to produce 3.3 grams of hydrogen gas, H₂, is approximately 18.82 grams of sodium metal

Explanation:

The given mass of hydrogen gas produced = 3.3 grams

The molar mass of hydrogen gas, H₂ = 2.016 g/mol

The number of moles of hydrogen gas in 3.3 grams of H₂, 'n', is given as follows;

n = Mass/(Molar mass)

n = 3.3 g/(2.016 g/mol) = 1.63690476 moles of H₂

The reaction of sodium and water can be written as follows;

2Na + 2H₂O → 2NaOH + H₂ (g)

2 moles of sodium produces 1 mole of hydrogen gas, H₂

Therefore;

1.63690476/2 moles of sodium will produce 1.63690476 moles of hydrogen gas, H₂

The molar mass of sodium, Na ≈ 22.989 g/mol

The mass of 1.63690476/2 moles of sodium, 'm', is given as follows;

m = 1.63690476/2 moles × 22.989 g/mol ≈ 18.8154018 grams ≈ 18.82 grams

The mass of sodium that will produce 3.3 grams of hydrogen, m ≈ 18.82 grams of sodium metal.

8 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 descr

ra1l [238]

Answer: 0.887 g of $MnO_2$ should be added to excess HCl(aq).

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 805 torr = 1.06 atm (760torr=1atm)

V = Volume of gas = 235 ml = 0.235 L

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $25^0C=(25+273)K=298K$

$n=\frac{PV}{RT}$

$n=\frac{1.06atm\times 0.235L}{0.0820 L atm/K mol\times 298K}=0.0102moles$

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

According to stoichiometry:

1 mole of chlorine is produced by = 1 mole of $MnO_2$

Thus 0.0102 moles of chlorine is produced by = $\frac{1}{1}\times 0.0102=0.0102$ moles of $MnO_2$

Mass of $MnO_2$ = $moles\times {\text {Molar mass}}=0.0102mol\times 87g/mol=0.887g$

0.887 g of $MnO_2$ should be added to excess HCl(aq).

7 0

3 years ago

A displacement reaction takes place when calcium is placed in a solution of zinc chloride. Which substances will most likely be

Karolina [17]

Hi!

The correct option would be A.

This is because the displacement reaction would take place as follows

Ca + ZnCl2 --> CaCl2 + Zn

A displacement reaction is one in which a substitution occurs, as the more reactive element in the mixture replaces one that is less reactive.

In the electrochemical series, we find Ca higher than Zn, which is indicative of Ca being more reactive, and having the capacity to displace Zn to form a compound.

Option D would be incorrect as no such substitution occurs.

Option B would be incorrect because again, there is no substitution occurring, and also because two metals alone (Ca and Zn in our case) can never react to form a compound.

Option C would be incorrect because it is not possible because CaCl and ZnCl are forms that are too unstable to exist due to an overall positive charge.

Hope this helps!

5 0

3 years ago

What did early scientists assume that the polar caps had in common?

Vilka [71]

Answer:

They assumed they both had water.

Explanation:

Because they only could look at it through telescopes that were not advanced

3 0

3 years ago

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

7 0

3 years ago

True or false? Freshwater can occur as a gas, liquid, or solid.