Ask question

Ask question

All categories

3 years ago

5

suppose that an ionic compound m represents a metal that could form more than one type of ion. in the formula MO the charge of t

he m ion would be?

1 answer:

blsea [12.9K]3 years ago

7 0

Answer:

Explanation:

You might be interested in

Use a chemical equation to show how aqueous hno3 fits the definition of an acid

Lunna [17]

An acid has several definitions one would be that it is a proton donor and an electron donor. Also, it said to produce an H+ ion when in solution. For nitric acid in solution, it dissociates into ions which are NO- ions and H+ ions. So, it must be an acid. It has a dissociation reaction:

HNO3 = H+ + NO3-

7 0

3 years ago

Why do you see lightning before you hear the thunder?

krok68 [10]

B,speedof light>speed of sound

8 0

2 years ago

Read 2 more answers

If 12g nitrogen gas,0.40 of H2 gas and 9.0 gram of oxygen are put into 1 litre container of. 27°C what is the total pressure on

Vadim26 [7]

Answer:

Total pressure = 27.35 atm

Explanation:

Given data:

Mass of nitrogen = 12 g

Mass of H₂ = 0.40 mol

Mass of oxygen = 9.0 g

Volume of Container = 1 L

Temperature = 27 °C (27+273 = 300 K)

Total Pressure = ?

Solution:

First of all we will calculate the number of moles of individual gas.

Number of moles = mass/ molar mass

Number of moles = 12 g/ 28 g/mol

Number of moles = 0.43 mol

Pressure of N₂:

PV = nRT

P = nRT/V

P = 0.43 mol × 0.0821 atm.L/mol.K× 300 K/ 1 L

P = 10.6 atm

Number of moles of Oxygen:

Number of moles = mass/ molar mass

Number of moles = 9 g/ 32 g/mol

Number of moles = 0.28 mol

Pressure of O₂:

PV = nRT

P = nRT/V

P = 0.28 mol × 0.0821 atm.L/mol.K× 300 K/ 1 L

P = 6.9 atm

Pressure of H₂:

PV = nRT

P = nRT/V

P = 0.40 mol × 0.0821 atm.L/mol.K× 300 K/ 1 L

P = 9.85 atm

Total pressure:

Total pressure = Pressure of H₂ + Pressure of O₂ + Pressure of N₂

Total pressure = 9.85 atm + 6.9 atm + 10.6 atm

Total pressure = 27.35 atm

8 0

3 years ago

Calculate the whole-number ratio of staggered to eclipsed conformers that are present at room temperature. Use the Actual G˚ bar

natka813 [3]

Answer:

The ratio of staggered to eclipsed conformers is 134

Explanation:

It is possible to determine the ratio of staggered to eclipsed conformers of a reactant, using the equilibrium:

Staggered ⇄ Eclipsed

Keq = [Eclipsed] / [Staggered]

That means Keq is equal to the ratio we need to find:

Using:

G˚= -RTln Keq

<em>Where G° = -12133.6J/mol</em>

<em>R is gas constant: 8.314J/molK</em>

<em>T is absolute temperature: 298K</em>

<em />

-12133.6J/mol= -8.314J/molK*298K ln Keq

4.8974 = ln Keq

134 = Keq = [Eclipsed] / [Staggered]

<h3>The ratio of staggered to eclipsed conformers is 134</h3>

3 0

2 years ago

Dahasolnce [82]

Volume of CO₂ obtained : 13 L

<h3>Further explanation</h3>

Reaction

C₂H₅OH + 3 O₂ ⇒ 2 CO₂ + 3 H₂0 + 8842 Joules

moles of ethanol=0.28

From equation, mol ratio ethanol : CO₂ = 1 : 2, so mol CO₂ :

$\tt \dfrac{2}{1}\times 0.28=0.56$

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

Then volume of CO₂ :

$\tt V~CO_2=mol\times 22.4=0.56\times 22.4=12.544\approx 13~L$

5 0

2 years ago