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Nostrana [21]
3 years ago
6

If the pressure of a gas is 2.43 atm what is it’s pressure in kpa

Chemistry
1 answer:
Viktor [21]3 years ago
8 0

The pressure of the gas is 246 kPa.

<em>p</em> = 2.43 atm × (101.325 kPa/1 atm) = 246 kPa  

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If 115grams of a substance reacts with 84grams of another substance, what will be the mass of the products after the reaction?
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Isn’t it 115+84 so 199
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What is bulimia? can someone explain it to me.
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3 years ago
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Can someone please give me a simple equation on the mass of an isotope
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Mass 1 + %abundance of first isotope + Mass 2 + %abundance of second isotope
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5 0
3 years ago
The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.
azamat

Answer:

(a) m_{gold}=7.322g

(b)

V_{gold}=0.379cm^3

V_{copper}=0.122cm^3

(c) \rho _{coin}=15.94g/cm^3

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:

m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3

m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3}  \\\\V_{copper}=0.122cm^3

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\  \\\rho _{coin}=15.94g/cm^3

Best regards.

3 0
3 years ago
Read 2 more answers
How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is -5,500 kJ/mol CsH18? C8H18 +
aalyn [17]

Answer : The correct option is, (D) 3600 kJ

Explanation :

Mass of octane = 75 g

Molar mass of octane = 114.23 g/mole

Enthalpy of combustion = -5500 kJ/mol

First we have to calculate the moles of octane.

\text{ Moles of octane}=\frac{\text{ Mass of octane}}{\text{ Molar mass of octane}}=\frac{75g}{114.23g/mole}=0.656moles

Now we have to calculate the heat released in the reaction.

As, 1 mole of octane released heat = -5500 kJ

So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)

                                                                   = -3608 kJ

                                                                   ≈ -3600 kJ

Therefore, the heat released in the reaction is 3600 kJ

5 0
2 years ago
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