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3 years ago

Which is a factor into Tyrone in the average atomic mass of an element

Chemistry

1 answer:

nikklg [1K]3 years ago

6 0

The mass numbers of the different isotopes of that element are averaged according to their respective abundances in nature.

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Convert 0.37 kilograms to grams.

finlep [7]

0.37 kilogramme is the same as 370 grams

5 0

4 years ago

Read 2 more answers

Calculate the molarity of the two solutions.

beks73 [17]

Answer:

A = Molarity = 0.22 M

B = Molarity = 0.36 M

Explanation:

Given data:

For first solution:

number of moles = 0.550 mol

Volume of solution = 2.50 L

Molarity = ?

Molarity:

Formula:

Molarity = number of moles of solute / volume of solution in L.

Molarity = 0.550 mol / 2.50 L

Molarity = 0.22 M

For second solution:

Mass of NaCl = 15.7 g

Volume of solution = 709 mL or 709/1000 = 0.709 L

Molarity = ?

Solution:

Number of moles = mass / molar mass

Number of moles = 14.7 g/ 58.44 g/mol

Number of moles = 0.252 mol

Molarity:

Molarity = number of moles of solute / volume of solution in L.

Molarity = 0.252 mol / 0.709 L

Molarity = 0.36 M

8 0

3 years ago

Which of the following is a physical model of the Sun? A. an equation that describes the Sun’s motion B. a chart that lists the

Bezzdna [24]

Answer:

It would be E. A small yellow ball that represents the Sun Reset Next i got the answer from somebody else

Explanation:

hope this helps :)

4 0

2 years ago

Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the followi

marta [7]

Answer : The value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

Explanation :

The following equilibrium reactions are :

(1) $2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2$ $K_1=5.40\times 10^{-16}$

(2) $2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l)$ $K_2=1.06\times 10^{10}$

(3) $CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g)$ $K_3=2.68\times 10^{-9}$

The final equilibrium reaction is :

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$ $K_{goal}=?$

Now we have to calculate the value of $K_{goal}$ for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

$K_{goal}=\sqrt{K_1\times K_2\times K_3}$

Now put all the given values in this expression, we get :

$K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}$

$K_{goal}=1.238\times 10^{-7}$

Therefore, the value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

3 0

3 years ago

Great amounts of atomic energy are released when a _______reaction occurs.

Mariulka [41]

Great amounts of atomic energy are released when a _______reaction occurs.

Great amounts of atomic energy are released when a chemical reaction occurs. The process can be an exothermic reaction or endothermic reaction depending on the substances involved in the reaction.

8 0

3 years ago