At STP or standard conditions, which list of elements contains a solid, a liquid, and a gas?

Chemistry

1 answer:

SIZIF [17.4K]3 years ago

4 0

Answer:

B.hafnium, mercury,helium

Explanation:

because the atomic number of hafnium is 72is a solid, mercury is a gas the atomic number is 80,

helium is a gas the atomic number is 2

You might be interested in

Estimate the standard internal energy of formation of liquid methyl acetate (methyl ethanoate, ch3cooch3) at 298 k from its stan

LenKa [72]

solution:

$the reaction for formation of methyl acetate is\\CH_{3}OH+CH_{3}COOH----------->CH_{3}COOCH_{3}+H_{2}O\\standard internal energy \\\delta u=\delta H -\delta(pv)=\delta H -\delta n\times RT\\where \delta n=change in number of moles=0\\\delta u=\delta H\\=-442kj/mol$

6 0

3 years ago

Read 2 more answers

Calculate the molarity of 29.9 g of MgS in 911 mL of solution??

LUCKY_DIMON [66]

0.58 is the molarity of 29.9 g of MgS in 911 mL of solution.

Define the molarity of a solution.

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

Given data:

Volume of solution = 911 mL=0.911L

Given the mass of MgS = 29.9 g

Molar mass of MgS = 56.38 g/mol

Moles of MgS = 0.5303299042

$Molality = \frac{Moles \;solute}{Volume \;of \;solution \;in \;litre}$

$Molality = \frac{0.5303299042}{0.911L}$

Molality = 0.5821403998=0.58

Hence, 0.58 is the molarity of 29.9 g of MgS in 911 mL of solution.

Learn more about Molarity here:

brainly.com/question/2817451

#SPJ1

3 0

2 years ago

1. Determine the final volume of a 1.5 M HCl solution prepared from 20.0 mL

nadezda [96]

Explanation:

Because molarity is classified as moles of solute per liter of water, dilution of the water may result in a reduction of its concentration.

Therefore, because the amount of moles of solute has to be constant for dilution, you will use the molarity and volume of that same target solution to calculate how many moles of solute will be present in the sample of the stock solution that you dilute.

c = $\frac{n}{v}$