How many protons, neutrons, and electrons are in a silver atom, with a mass number of 108?

Alex

Answer:

space biology is a fundamental component of space life science. it focuses on smaller organisms such as cell, animals and plants.

4 0

3 years ago

What percentage of boys Chose walking

mafiozo [28]

It would of been nice if you were to post your question so others could actually answer it.

6 0

3 years ago

What is the total pressure of air in lungs of an individual with oxygen at 100. mmHg, nitrogen at 573 mmHg, carbon dioxide at 0.

ryzh [129]

Answer: The total pressure of air in lungs of an individual is 760.28 mm Hg

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_A+p_B+p_C...$

Given : $p_{total}$ =total pressure of gases = ?

$p_{O_2}$ = partial pressure of oxygen = 100 mm Hg

$p_{N_2}$ = partial pressure of nitrogen = 573 mm Hg

$p_{CO_2}$ = partial pressure of Carbon dioxide = 0.053 atm = 40.28 mm Hg(1 atm = 760 mmHg)

$p_{H_2O}$ = partial pressure of water vapor = 47 torr = 47 mm Hg (1torr=1 mm Hg)

putting in the values we get:

$p_{total}=(100+573+40.28+47)mmHg$

$p_{total}=760.28mmHg$

Thus the total pressure of air in lungs of an individual is 760.28 mm Hg

5 0

3 years ago

Jwjajajahwhab a a skwiwiwiahah

FrozenT [24]

Thank you :))) for the points have a great day

8 0

3 years ago

Read 2 more answers

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

scZoUnD [109]

Answer: The mass percent of potassium bromide in the mixture is 9.996%

Explanation:

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

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

For lead (II) bromide:

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