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MariettaO [177]

2 years ago

8

10. Which best explains why the total mass of the product(s) would be less than the total weight of the

1 answer:

Olegator [25]2 years ago

4 0

Answer:

D. Gases were released to the atmosphere

Explanation:

In accordance to the law of conservation of mass, the total amount of reactants must equate the total amount of products at the end of the reaction because matter can not be lost or created. However, certain changes like gas evolution, formation of precipitate etc. indicates the occurrence of a chemical reaction.

In a chemical reaction, the total mass of the product(s) would be less than the total weight of the reactant(s) because GASES, which constituted part of the mass of the reaction, WERE RELEASED INTO THE ATMOSPHERE. However, if the mass of the gas released can be accounted for, the amount of reactants and products must balance.

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How many moles of magnesium, Mg, are there in 4.75 grams of magnesium?

Nina [5.8K]

Molar mass Mg = 24.3 g/mol

1 mole mg ------------ 24.3 g
?? moles mg --------- 4.75 g

4.75 x 1 / 24.3 => 0.195 moles of Mg

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2 years ago

AysviL [449]

Answer:

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3 years ago

HELPPP LOOK AT PICTURE

lina2011 [118]

Answer:

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Explanation:

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5 0

2 years ago

How many moles of HNO3 are present if 4.90×10−2 mol of Ba(OH)2 was needed to neutralize the acid solution?

Amiraneli [1.4K]

Answer:

0.098 moles

Explanation:

Let y represent the number of moles present

1 mole of Ba(OH)₂ contains 2 moles of OH- ions.

Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.

To get the y moles, we then do cross multiplication

1 mole * y mole = 2 moles * 0.049 mole

y mole = 2 * 0.049 / 1

y mole = 0.098 moles of OH- ions.

1 mole of OH- can neutralize 1 mole of H+

Therefore, 0.098 moles of HNO₃ are present.

3 0

2 years ago

For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [

sveta [45]

Explanation:

The given data is as follows.

$E^{o}$ = 0.483, $[Co^{3+}]$ = 0.173 M,

$[Co^{2+}]$ = 0.433 M, $[Cl^{-}]$ = 0.306 M,

$P_{Cl_{2}}$ = 9.0 atm

According to the ideal gas equation, PV = nRT

or, P = $\frac{n}{V}RT$

Also, we know that

Density = $\frac{mass}{volume}$

So, P = MRT

and, M = $\frac{P}{RT}$

= $\frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}$

= $\frac{9.0}{24.436}$

= 0.368 mol/L

Now, we will calculate the cell potential as follows.

E = $E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}$

= $0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}$

= $0.483 - 0.02955 log \frac{0.0689}{0.0162}$

= $0.483 - 0.02955 \times 0.628$

= 0.483 - 0.0185

= 0.4645 V

Thus, we can conclude that the cell potential of given cell at $25^{o}C$ is 0.4645 V.

4 0

3 years ago