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puteri [66]
3 years ago
14

What is the volume of 0.5 moles of a gas at 2 atmospheres of pressure and 15°c​

Chemistry
1 answer:
liq [111]3 years ago
4 0

Answer:

6l

Explanation: convert temperature to kelvin by adding 273 and then input the values into the formula with the given constant

2*v=0.5*0.8206*288 then divide both sides by 2 and get the amount in litres which is 6

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Which planet is the most massive? A.Jupiter B.Mercury C.Venus D.Neptune
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Jupiter is the biggest planet in are solar system

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3 years ago
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Use the Henderson-Hasselbalch equation, eq. (3), to calculate the pH expected for a buffer solution prepared from this acid and
notka56 [123]

Answer:

pH=4.56

Explanation:

Hello there!

In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol

And the concentrations are:

[acid]=0.000855mol/0.025L=0.0342M

[base]=0.000781mol/0.025L=0.0312M

Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

pH=-log(2.5x10^{-5})+log(\frac{0.0312M}{0.0342M} )\\\\pH=4.60-0.04\\\\pH=4.56

Best regards!

6 0
2 years ago
Which answer provides the correct name for the following hydrocarbon?
ivanzaharov [21]
1) Hydrocarbon: CH3 - CH2 - CH2 - CH2 - CH3

2) Only single bonds => alkane => sufix ane

3) no substitutions

4) 5 carbons = > prefix penta.

Therefore, the name is pentane.
4 0
3 years ago
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Calculate the volume of dry co2 produced at body temperature (37 ∘c) and 0.970 atm when 25.5 g of glucose is consumed in this re
katrin2010 [14]

I believe the balanced chemical equation is:

C6H12O6 (aq) + 6O2(g) ------> 6CO2(g) + 6H2O(l) 

 

First calculate the moles of CO2 produced:

moles CO2 = 25.5 g C6H12O6 * (1 mol C6H12O6 / 180.15 g) * (6 mol CO2 / 1 mol C6H12O6)

moles CO2 = 0.8493 mol

 

Using PV = nRT from the ideal gas law:

<span>V = nRT  / P</span>

V = 0.8493 mol * 0.08205746 L atm / mol K * (37 + 273.15 K) / 0.970 atm

<span>V = 22.28 L</span>

6 0
3 years ago
What is the boiling point of a solution of 76 g of water dissolved in 500 mL of acetic acid, CH3COOH?
tatuchka [14]

Answer:

127.3° C, (This is not a choice)

Explanation:

This is about the colligative property of boiling point.

ΔT = Kb . m . i

Where:

ΔT = T° boling of solution - T° boiling of pure solvent

Kb = Boiling constant

m = molal (mol/kg)

i = Van't Hoff factor (number of particles dissolved in solution)

Water is not a ionic compound, but we assume that i = 2

H₂O →  H⁺  +  OH⁻

T° boling of solution - 118.1°C =  0.52°C . m . 2

Mass of solvent =  Solvent volume / Solvent density

Mass of solvent = 500 mL / 1.049g/mL → 476.6 g

Mol of water are mass / molar mass

76 g / 18g/m = 4.22 moles

These moles are in 476.6 g

Mol / kg = molal → 4.22 m / 0.4766 kg = 8.85 m

T° boling of solution =  0.52°C . 8.85 m . 2 + 118.1°C =  127.3°C

6 0
3 years ago
Read 2 more answers
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