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2 years ago

Identify the number of significant figures in each value. 147,320 g

Chemistry

2 answers:

podryga [215]2 years ago

5 0

There is 5 significant figures in this value. The zero at the end is not considered significant.

Sidana [21]2 years ago

4 0

There are 5 significant figures in 147,320.

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Phosphorus pentachloride reacts with water to form hydrochloric acid and phosphoric acid. How many total moles of acid are forme

klio [65]

Answer:

0,13 moles of acid are produced

Explanation:

The reaction of the problem is:

PCl₅ + 4H₂O → 5HCl + H₃PO₄

Based on the reaction, 1 mole of PCl₅ produces 6 moles of acid (5 moles of HCl + 1 mole of H₃PO₄).

The molecular mass of PCl₅ is:

1P = 30,97g/mol + 5Cl = 5×35,45g/mol = 208,24 g/mol

That means 4,5g of PCl₅ are:

4,5g PCl₅×(1mol / 208,24g) = 0,0216 moles of PCl₅. As 1 mole of PCl₅ produces 6 moles of acid, 0,0216 moles of PCl₅ produce:

0,0216 moles PCl₅ × (6 moles acid / mole of PCl₅) =

0,13 moles of acid are produced

I hope it helps!

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3 years ago

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protones

Explanation:

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3 years ago

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Which law states that at constant temperature the pressure of a given amount of gas is inversely proportional to the volume of t

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That law is known as Boyle's Law, "The volume of a given mass of a gas is inversely related to pressure when the temperature is constant"

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3 years ago

The salts NaCl and CaCl

Irina18 [472]

The answer is 8 :)
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3 years ago

For the following electrochemical reaction: Al3+(aq) + 3e -> Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -> 2F (aq) Calculate

makvit [3.9K]

Answer: The standard electrode potential of the cell is 4.53 V.

Explanation:

We are given:

$E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=2.87-(-1.66)=4.53V$

Hence, the standard electrode potential of the cell is 4.53 V.

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2 years ago