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3 years ago

7

Suppose 0.274 mol of a gas occupy 6.62 l at a given temperature and pressure. what volume will 0.878 mol occupy at the same temp

erature and pressure?

1 answer:

Taya2010 [7]3 years ago

6 0

The volume that 0.878 mol occupy at the same temperature and pressure is calculated as follows

if 0.274 mol occupy 6.62 L what about 0.878 mol

by cross multiplication

=0.878 x6.62/0.274=21.21 L

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The ion p3− has _____ protons and _____ electrons. enter your answers as integers separated by a comma.'

katrin2010 [14]

The symbol $P$ is for phosphorus having atomic number 15. That means it has $15 protons and 15 electrons$ in its neutral state.

The given symbol $P^{3-}$ has a -3 charge that means it has gained 3 electrons so, the number of electrons in the $P^{3-}$ is: $15+3 = 18$ . The number of protons will remain same that is the number of protons = 15.

So, the ion $P^{3-}$ has $15 protons and 18 electrons$ that gives $15-18=-3$ charge on the $P^{3-}$ ion.

So, the answer is $15 , 18$ .

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3 years ago

Question List (4 items) (Drag and drop into the appropriate area) Find the volume of HCl that will neutralize the base. Find the

expeople1 [14]

The question is incomplete, the complete question is:

The solubility of slaked lime, $Ca(OH)_2$ , in water is 0.185 g/100 ml. You will need to calculate the volume of $2.50\times 10^{-3}$ M HCl needed to neutralize 14.5 mL of a saturated

<u>Answer:</u> The volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

<u>Explanation:</u>

Given values:

Solubility of $Ca(OH)_2$ = 0.185 g/100 mL

Volume of $Ca(OH)_2$ = 14.5 mL

Using unitary method:

In 100 mL, the mass of $Ca(OH)_2$ present is 0.185 g

So, in 14.5mL. the mass of $Ca(OH)_2$ present will be = $\frac{0.185}{100}\times 14.5=0.0268g$

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

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

Given mass of $Ca(OH)_2$ = 0.0268 g

Molar mass of $Ca(OH)_2$ = 74 g/mol

Plugging values in equation 1:

$\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol$

Moles of $OH^-$ present = $(2\times 0.000362)=0.000724mol$

The chemical equation for the neutralization of calcium hydroxide and HCl follows:

$Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O$

By the stoichiometry of the reaction:

Moles of $OH^-$ = Moles of $H^+$ = 0.000724 mol

The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(2)

Moles of HCl = 0.000724 mol

Molarity of HCl = $2.50\times 10^{-3}$

Putting values in equation 2, we get:

$2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL$

Hence, the volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

5 0

3 years ago

Convert 1.25 atm to bar.

Tasya [4]

Answer: 1.27 bar

Explanation:

1 atm = 1.01325 bar

1.25 atm = Z (let Z be the unknown value)

To get the value of Z, cross multiply

Z x 1 atm = 1.25 atm x 1.01325 bar

1 atm•Z = 1.2665625 atm•bar

To get the value of Z, divide both sides by 1 atm

1 atm•Z/1 atm = 1.2665625 atm•bar/1atm

Z = 1.2665625 bar

(Round up Z to the nearest hundredth as 1.27 bar)

Thus, 1.25 atm when coverted gives 1.27 bar

8 0

3 years ago

How many moles of KNO3 are in 500 mL of 2.0 M KNO3?<br> ___ mol KNO3

weeeeeb [17]

Answer: 1mole

Explanation:

Mole = concentration× Volume (dm3)

Mole = 2× 500/1000

4 0

3 years ago

Read 2 more answers

One of the reactions in a blast furnace used to reduce iron is shown above. How many grams of Fe2O3 are required to produce 15.5

Salsk061 [2.6K]

Answer : The correct option is, (b) 22.1 g

Solution : Given,

Mass of iron = 15.5 g

Molar mass of iron = 56 g/mole

Molar mass of $Fe_2O_3$ = 160 g/mole

First we have to calculate the moles of iron.

$\text{Moles of Fe}=\frac{\text{Mass of Fe}}{\text{Molar mass of Fe}}=\frac{15.5g}{56g/mole}=0.276moles$

Now we have to calculate the moles of $Fe_2O_3$ .

The balanced reaction is,

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

From the balanced reaction, we conclude that

As, 2 moles of iron obtained from 1 mole of $Fe_2O_3$

So, 0.276 moles of iron obtained from $\frac{0.276}{2}=0.138$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{Mass of }Fe_2O_3=\text{Moles of }Fe_2O_3\times \text{Molar mass of }Fe_2O_3$

$\text{Mass of }Fe_2O_3=(0.138mole)\times (160g/mole)=22.08g=22.1g$

Therefore, the amount of $Fe_2O_3$ required are, 22.1 grams.

6 0

3 years ago