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2 years ago

What is the molarity of an h3po4 solution if 15.0 ml is completely neutralized by 38.5 ml of 0.150m naoh?

Chemistry

2 answers:

polet [3.4K]2 years ago

6 0

The correct answer is 0.128M H3PO4

just olya [345]2 years ago

4 0

The molarity of H3PO4 solution if 15.0 ml is completely neutralized by 38.5 ml of 0.15m naoh is calculated as follows
find moles of NaOH used = molarity x volume
= 38.5 x 0.15 = 5.775 moles

write the reacting equation
3NaOH + H3PO4 = Na3PO4 + 3H2O

from the equation the reacting ratio between NaOH to H3Po4 which is 3:1 the moles of H3PO4 is therefore = 5.775/3 = 1.925 moles

molarity of H3PO4 is therefore = moles /volume
= 1.925/15 = 0.128 M

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Air in a 0.3 m3 cylinder is initially at a pressure of 10 bar and a temperature of 330K. The cylinder is to be emptied by openin

sergiy2304 [10]

Answer:

(a) Temperature = 330 K, and mass = 0.321 kg

(b) T₂ = 171.56 K, mass = 0.32223 kg

Explanation:

For a constant temperature process we have

p₁v₁ = p₂v₂

Where p₁ = initial pressure = 10 bar = 1000000 Pa

p₂ = final pressure = 1 atm = 101325 Pa

v₁ initial volume = 0.3 m³

v₂ = final volume = unknown

From the relation we have v₂ = 2.96 m³

Therefore at constant temperature 2.93 m³ - 0.3 m³ or 2.66 m³ will be expelled from the container

Temperature = 330 K, and mass =

Also from the relation p1v1 = mRT1

We have, (1000000×0.3)/(8314×330) = 109..337 mole

For air mass

Mass = 3.171 kg

After opening we have

p2v2/(RT1) = n2 = 11.07 mol or 0.321 kg

(b) This is said to be adiabatic condition hence

Here

But cp = 29 (J/mol K).

and p₁v₁ = RT₁ therefore R = 1000000*0.3/330 = 909.1 J/mol·K

And For perfect gas γ = 1.4

Hence T₂ = 171.56 K

γ =cp/cv therefore cv=cp/γ = 29/1.4 = 20.714 (J/mol K). and R =cp-cv = 8.29 J/mol·K

Therefore p1v1/(RT1) = 109.66 moles and we have

p2v2/(R×T2) = 11.11 mole left

For air that is 0.32223 kg

5 0

2 years ago

A metal oxide with the formula mo contains 15.44% oxygen. in the box below, type the symbol for the element represented by m.

lana66690 [7]

Answer: The element represented by M is Strontium.

Explanation:

Let us consider the molar mass of metal be 'x'.

The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol

It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:

$\frac{15.44}{100}\times (x+16)=16g/mol\\\\(x+16)=\frac{16g/mol\times 100}{15.44}\\\\x=(103.626-16)g/mol\\\\x=87.62g/mol$

The metal atom having molar mass as 87.62/mol is Strontium.

Hence, the element represented by M is Strontium.

8 0

1 year ago

Read 2 more answers

Automobiles are often implicated as contributors to global warming because they are a source of the greenhouse gas CO2. How many

kondor19780726 [428]

Answer:

5,747.82 pounds of $CO_2$ would your car release in a year if it was driven 170 miles per week.

Explanation:

Mileage of the car = 27.5 miles/gal

Distance driven by car in week = 170 miles

Distance driven by car in a day= $\frac{170 miles}{7}=24.286 mile$

Volume of gasoline used in a day : V

$\frac{24.286 mile}{27.5 miles/gal}=0.8831 gal=0.8831\times 3785.41 cm^3=3,342.96 cm^3$

$1 gal = 3785.41 mL = 3785.41 cm^3$

Density of the gasoline = d= $0.692 g/cm^3$

Mass of the gasoline used in a day = m

$m=d\times V=0.692 g/cm^3\times 3,342.96 cm^3=2,313.33 g$

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

Moles of gasoline ( isooctane):

= $\frac{2,313.33 g}{114 g/mol}=20.292 mol$

According to reaction , 2 moles of isooctane gives 16 moles of carbon dioxide gas.Then 20.292 mole isooctane will give :

$\frac{16}{2}\times 20.292 mol=162.340 mol$ of carbon dioxide gas

Mass of 162.340 moles of carbon dioxide gas :

162.340 mol × 44 g/mol = 7,142.91 g

Mass of carbon dioxide gas produced in a day = 7,142.91 g

1 year = 365 days

Mass of carbon dioxide gas produced in a 365 days:

= 365 × 7,142.91 g = 2,607,163.494 g

1 pound = 453.592 g

$2,607,163.494 g=\frac{ 2,607,163.494}{453.592} pounds=5,747.82 pounds$

5,747.82 pounds of $CO_2$ would your car release in a year if it was driven 170 miles per week.

7 0

1 year ago

What are the relative numbers of h3o+ and oh- ions in an acidic and alkaline and a neutral solution?

lana [24]

Acid more H3O+ than OH-
Base less H3O+ than OH-

3 0

1 year ago

Read 2 more answers

Ancient Egyptians used a variety of lead compounds as white pigments in their cosmetics, including PbS, PbCO3, PbCl(OH), and Pb2

ser-zykov [4K]

Answer : The compound contains the highest percentage of lead (by mass) is, PbS.

Explanation :

To calculate the percentage of lead in sample, we use the equation:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100$

For $PbS$ :

Mass of $PbS$ = 239.3 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{239.3g}\times 100=86.58\%$

The percentage of lead in the $PbS$ is 86.58 %.

For $PbCO_3$ :

Mass of $PbCO_3$ = 267.2 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{267.2g}\times 100=77.55\%$

The percentage of lead in the $PbCO_3$ is 77.55 %.

For $PbCl(OH)$ :

Mass of $PbCl(OH)$ = 259.7 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{259.7g}\times 100=79.78\%$

The percentage of lead in the $PbCl(OH)$ is 79.78 %.

For $Pb_2Cl_2CO_3$ :

Mass of $Pb_2Cl_2CO3_$ = 545.3 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{545.3g}\times 100=37.99\%$

The percentage of lead in the $Pb_2Cl_2CO3_$ is 37.99 %.

Hence, from this we conclude that, the compound PbS contains the highest percentage of lead (by mass).

6 0

1 year ago