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Lemur [1.5K]
3 years ago
5

A 12.2-g sample of X reacts with a sample of Y to form 78.9g of XY. what mass of Y reacted?​

Chemistry
1 answer:
barxatty [35]3 years ago
3 0

Answer:

The answer to your question is:  y = 66.7g

Explanation:

To solve this problem we must remember the Lavoisier law of conservation of mass, that stated that the mass can not be created or destroyed but it changes forms.

So, the mass of the reactants is equal to the mass of the products.

Process

                x          +          y             ⇒           xy

             12.2g                   ?                         78.9 g

           12.2 g + y = 78.9g

           y = 78.9 - 12.2

           y = 66.7g

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At 0 degrees Celsius, a gas occupies 22.4L. How hot must the gas be in celcius to reach a volume of 25.0L
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Answer:

31.7 °C

Explanation:

Charles law states that for volume of a gas is directly proportional to the absolute temperature for a fixed amount of gas at constant pressure

we can use the following equation

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V2 is volume and T2 is temperature at second instance

temperature should be in kelvin scale

T1 - 0 °C + 273 = 273 K

substituting the values in the equation

22.4 L / 273 K = 25.0 L / T2

T2 = 304.7 K

temperature in celcius is - 304.7 K - 273 = 31.7 °C

the gas must be 31.7 °C to reach a volume of 25.0 L

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Calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. express the nu
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3 years ago
A 0.533 g sample of a carbonate salt is added to a flask. The only thing you know is that per formula unit, there is only one ca
dimulka [17.4K]

Answer:

\mathbf{CrCO_3}

Explanation:

The total mole of H_2SO_4 added = 15 × 1 × 10⁻³

= 15 × 10⁻³ mole

= 15 mmoles

the number of moles of NaOH added in order to neutralize the excess acid = 0.5905 ×  29.393 = 17.36 mmoles

the equation for the reaction is:

2NaOH + H_2SO_4 --------> Na_2SO_4 +2H_2O

i.e

2 moles of NaOH react with H_2SO_4

1 moles of  NaOH will react with 1/2  H_2SO_4

17.36 mmoles of NaOH = 1/2 × 17.36 mmoles of H_2SO_4

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Number of moles of H_2SO_4 that react with MCO₃ = Total moles of H_2SO_4 added - moles of H_2SO_4 reacted with NaOH

= (15 - 8.68) mmoles

= 6.32 mmoles

H_2SO_4 + MCO_3  ------>   M_2SO_4   +    H_2O   +   CO_2

1 mole of H_2SO_4 react with 1 mole of MCO_3

6.32 mmoles of H_2SO_4 = 6.32 mmoles of MCO_3

number of moles of MCO_3 = 6.32 10⁻³ moles

mass of  MCO_3 (carbonate salt) = 0.533 g

molar mass of MCO_3  = (M+60)g/mol

We all know that ;

number of moles = mass/molar mass

Then:

6.32 10⁻³ = 0.533 / (M+60)

(M+60) = 0.533/ 6.32 10⁻³

M + 60 = 84.34

M = 84.34 - 60

M = 24.34

Thus the element with the atomic mass of 24.34 is Chromium

The chemical formula for the compound is :  \mathbf{CrCO_3}

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