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sergij07 [2.7K]
3 years ago
13

What are all four Quantum numbers of 11th electron of magnesium​

Chemistry
1 answer:
ruslelena [56]3 years ago
3 0

Answer:n=0

l=0

ml=0

ms = =1/2 or +1/2

Explanation:

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a chemist adds of a calcium bromide solution to a reaction flask. calculate the mass in grams of calcium bromide the chemist has
mestny [16]

The mass of Calcium bromide added in the flask is 29.7 g.

<h3>What is Molarity? </h3>

Molarity is defined as the ratio of number of moles od solute to the number of volume of solution in litres.

Molarity = number of moles/ volume

<h3>Calculation of Moles</h3>

Number of moles = Molarity × volume

Given,

Molarity of Calcium bromide = 0.363 M

Volume of Calcium bromide = 410 mL

= 0.410L

By substituting all the value, we get

Number of moles = 0.363 × 0.410

= 0.148 mol

As we know that,

Molar mass of Calcium bromide = 199.89 g

<h3>What is Mole? </h3>

Mole is defined as the given mass of substance to the molar mass of substance.

Given mass = Moles × Molar mass

= 0.148 × 199.89

= 29.75 g

= 29.7 g (significant digit)

Thus, we calculated that the mass of Calcium bromide added in the flask is 29.7 g.

learn more about Molarity:

brainly.com/question/19517011

#SPJ4

DISCLAIMER:

The above question is incomplete. Below is the complete question

A chemist adds 410.0mL of a 0.363 M calcium bromide solution to a reaction flask. calculate the mass in grams of calcium bromide the chemist has added to the flask. round your answer to 3 significant digits.

4 0
2 years ago
How many oxygen atoms are in 1.40 g of quartz?
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Polonium has a mass number of 212 and an atomic number of 84. In a balanced nuclear reaction, polonium releases an alpha particl
antoniya [11.8K]

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D. 82

Explanation:

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Answer:

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Explanation:

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3 years ago
Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.2 g o
stich3 [128]

Answer: 0.0 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of butane

\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles

b) moles of oxygen

\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

According to stoichiometry :

2 moles of butane require 13 moles of O_2

Thus 0.09 moles of butane will require =\frac{13}{2}\times 0.09=0.585moles  of O_2

Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.

Thus all the butane will be consumed and 0.0 grams of butane will be left.

7 0
2 years ago
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