1) What relationship is needed to perform mole-mass conversion?

Chemistry

2 answers:

likoan [24]3 years ago

8 0

Answer:

$n = \frac{m}{mole mass}$

mixer [17]3 years ago

7 0

Answer:

Explanation:

A substance's molar mass is calculated by multiplying its relative atomic mass by the molar mass constant (1 g/mol). The molar mass constant can be used to convert mass to moles. By multiplying a given mass by the molar mass, the amount of moles of the substance can be calculated.

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The solid form of a substance is usually more dense than its liquid and gaseous forms. Similarly the liquid form is usually more

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2 years ago

10) In order to make spaghetti cook faster, a chef adds salt to water. How many moles of salt would he need to add to 1.0 kg wat

Vadim26 [7]

The answer is 4.9 moles.
Solution:
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Δt = i Kb m
we can rearrange the expression to solve for the molality m of the solution:
m = Δt / i Kb

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m = (105°C - 100°C) / (2 * 0.512 °C·kg/mol)
= 4.883 mol/kg

From the molality m of the solution of salt added in a kilogram of water, we can now find the number of moles of salt:
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3 years ago

Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container

Katarina [22]

Answer:

The average pressure in the container due to these 75 gas molecules is $P=9.72 \times 10^{-16} Pa$

Explanation:

Here Pressure in a container is given as

$P=\frac{1}{3} \rho$

Here

P is the pressure which is to be calculated

ρ is the density of the gas which is to be calculated as below

$\rho =\frac{mass}{Volume of container}$

Here

mass is to be calculated for 75 gas phase molecules as

$m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times 10^{-21} g$

Volume of container is 0.5 lts

So density is given as

$\rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3$