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Semmy [17]
2 years ago
8

What is the molar mass of AINO3? Explanation​

Chemistry
1 answer:
Verdich [7]2 years ago
5 0

Answer:

AINO3 - 89 g/mol

Explanation:

Molar mass of a compound or molecule is obtained by summing the individual atomic masses of the elements that makes the compound.

In this question, the elements  are Al, N and O

The molar mass is given as;

Molar mass = Al + N + 3 (O)

Al = 27

N = 14

O = 16

Molar mass = 27 + 14 + 3(16)

Molar mass = 89 g/mol

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Bombardment of cobalt-59 with a neutron produces a manganese-56 atom and another particle. What is this particle?A. an alpha par
Aleksandr-060686 [28]

Answer:A

Explanation:

A nuclear reaction is balanced by ensuring that the Masses and charges of te reactants and products are exactly balanced on the left and right hand side of the reaction equation. If there are 60 mass units on the LHS and manganese has only 56 mass units then four mass units are left. If there is no charge on a neutron and there is a charge of 27 on the cobalt, then two charges are left. Four mass units and a charge of +2 corresponds to a helium which is actually an alpha particle.

4 0
3 years ago
Use the table on the right to calculate each required quantity.
k0ka [10]

a. 35.8 KJ

b. 871 g

c. Fe

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3 0
3 years ago
20.00 mL of a 0.077 M solution of silver nitrate, AgNO3
melomori [17]
N= m/v n=0.077/.200 = 0.385
8 0
3 years ago
Draw the condensed structural formulas for all the possible haloalkane isomers that have four carbon atoms and a bromine.
meriva

<u>Answer:</u> The isomers are shown in the image below.

<u>Explanation:</u>

Isomers are defined as the chemical compounds having the same number and kinds of atoms but arrangement are different.

For the alkane having four carbon atoms and 1 bromine atom, the IUPAC name of the haloalkane is bromobutane

There are 4 possible isomers for the given haloalkane compound:

  1. 1-bromobutane
  2. 2-bromobutane
  3. 1-bromo-2-methylpropane
  4. 2-bromo-2-methylpropane

The isomers of the given organic compound is shown in the image below.

4 0
3 years ago
Question 15 How many grams of NaCl are required to make 500.0 mL of a 1.500 M solution? 58.40 g 175.3 g 14.60 g 43.83 g
ExtremeBDS [4]
Hi!

To make 500 mL of a 1,500 M solution of NaCl you'll require 43,83 g

To calculate that, you will need to use a conversion factor to go from the volume of the 1,500 M solution to the required grams. For this conversion factor, you'll use the definition for Molar concentration (M=mol/L) and the molar mass of NaCl. The conversion factor is shown below:

gNaCl=500mLsol* \frac{1L}{1000 mL}* \frac{1,500 mol NaCl}{1Lsol}* \frac{58,4428 g NaCl}{1 mol NaCl} \\ =43,83gNaCl

Have a nice day!
4 0
3 years ago
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