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pshichka [43]
3 years ago
14

What is the symbol of the element that is classified as an alkali metal and is in period 4?

Chemistry
1 answer:
djyliett [7]3 years ago
3 0

Answer:

Potassium (K) [First element in period 4]

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A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out
scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol

  • The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = \frac{2}{1}\times 0.0021=0.0042mol of potassium bromide

  • Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g

  • To calculate the percentage composition of KBr in the mixture, we use the equation:

\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0
3 years ago
What is one result of a chemical change?
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A new material is formed in <span>result of a chemical change. Typically, the chemical changes always make the new material.</span><span />
3 0
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Number of elments in hcooh
igomit [66]
3 elements are present in HCOOH - hydrogen, oxygen, and carbon.
7 0
3 years ago
How many moles of N2O5 are needed<br> to produce 7.90 g of NO2?
Roman55 [17]

Answer:

0.085 moles of  N₂O₅ are needed

Explanation:

Given data:

Mass of NO₂ produces = 7.90 g

Moles of N₂O₅ needed = ?

Solution:

2N₂O₅       →     4NO₂  + O₂

Number of moles of NO₂ produced :

Number of moles = mass/ molar mass

Number of moles = 7.90 g/ 46 g/mol

Number of moles = 0.17 mol

now we will compare the moles of NO₂   with N₂O₅.

                NO₂          :       N₂O₅

                  4            :          2

                0.17          :         2/4×0.17 = 0.085 mol

Thus, 0.085 moles of  N₂O₅ are needed.

4 0
3 years ago
Why is an elements atomic mass not listed as a whole number in the periodic table
FinnZ [79.3K]

Answer:

They aren't listed as a whole number, because an atom's mass is not always a whole number. The mass differs between types of atoms as well.

7 0
3 years ago
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