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Vlada [557]
3 years ago
9

A student has a sample of isopropanol (C3H,OH) that has a mass of 78.6 g. The molar mass of isopropanol is 60.1 g/mol. How

Chemistry
2 answers:
Sauron [17]3 years ago
4 0
I think it goes like this.

mass = 78.6 g
molar mass = 60.1 g/mol
amount (in mols) = mass/mol
78.6 g / 60.1 g/mol = 1.30782 mol

^^rounded to 3 sig figs, final answer = 1.31mol
lisov135 [29]3 years ago
4 0

Answer:

\boxed{\text{ 1.308 mol}}

Explanation:

\text{Moles} = \text{78.6 g } \times \dfrac{\text{1 mol}}{\text{60.1 g}} = \text{1.308 mol}\\\\\text{ The sample contains $\boxed{\textbf{ 1.308 mol}}$ of isopropyl alcohol}

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an element has atomic number 10 and an atomic mass of 20. how many neutrons are in the atom of this element
Flura [38]
Hey there!

n = A - Z

n =  20 - 10

n = 10 
6 0
3 years ago
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A piston contains 5.00 L of gas at a pressure of 101 kPa. The piston is compressed until the gas occupies 2.00 L of space. What
galben [10]

Answer: 252.5 kPa

Explanation:

Given that:

initial volume of gas V1 = 5.00 L

initial pressure of gas P1 = 101 kPa.

new Volume V2 = 2.00 L

new pressure P2 = ?

Since, only pressure and volume are involved, apply the formula for Boyle's law

P1V1 = P2V2

101 kPa x 5.0L = P2 x 2.00L

505 = P2 x 2.00L

P2 = 505/2.00

P2 = 252.5 kPa

Thus, the new pressure of gas inside the piston is 252.5 kPa

4 0
3 years ago
a 1 l container contains 02 gas at stp. If the diameter of the o2 molecule is 3.75 * 10^-10 m the mean free path of the o2 molec
Temka [501]

Answer:

ion know man

Explanation:

just wanted the points

5 0
3 years ago
What volume of Ar at 53.6 °C and 1.31 atm contains the same number of particles as 0.584 L of H2 at 19.6 °C and 2.10 atm?
solniwko [45]

Answer:

Volume = 1.043L

Explanation:

First off, we have to calculate the number of particles H2 at that condition have.

We were given the following;

Volume (V) = 0.584L

Temperature = 19.6°C + 273 = 292.6 K (Converting to Kelvin Temperature)

Pressure = 2.10 atm

Using the ideal gas equation;

PV = nRT,

where R = gas constant = 0.0821 L atm K−1 mol−1

making n subject of formular;

n = PV / RT

n = (2.1 * 0.584) / (0.0821 * 292.6)

n = 1.2264 / 24.02246

n = 0.05098 moles

Avogadro’s number is defined as the number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance. It is equal to 6.022×1023 mol-1

Number of particles present in 0.05098 moles of H2 is given as;

1 mole = 6.022×1023

0.05098 = x

Upon cross multiplication, we have;

x = 0.05098 * 6.022×1023 = 0.307×1023 = 3.07×1022 particles.

equal number of moles would have the same number of particles. So we just have to find the volume when the number of moles of Ar is 0.05098.

We were given the following for Ar;

Volume (V) = ?

Temperature = 53.6°C + 273 = 326.6 K (Converting to Kelvin Temperature)

Pressure = 1.31 atm

Using the ideal gas equation;

PV = nRT,

where R = gas constant = 0.0821 L atm K−1 mol−1

making V subject of formular;

V = nRT / P

V = ( 0.05098 * 0.0821 * 326.6) /  1.31

V = 1.043L

5 0
3 years ago
A student titrates an unknown amount of potassium hydrogen phthalate (khc8h4o4, abbreviated khp) with 20.46 ml of 0.1000 m naoh
Serga [27]
<span>0.1000 M NaOH means 0.1 mole per litre so 20.46ml has 20.46/1000 x 0.1 = 0.002046 moles - now this many moles of KHP is 0.002046 x 204.22=0.4178 g KHP</span>
3 0
3 years ago
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