Chemistry: How many grams are there in 3 mol NaNO3

How many grams are there in 3 mol NaNO3, №13522654, 22.07.2020 19:26

Mathematics

100 rules: have to answer correctly have to answer all 16 questions only answer if you only know the answer partial answers or incomplete answers not accepted will be reported point stealers beware 1) how many moles are there in 24.5 grams of cof3? 2) how many molecules are there in 150.00 grams of na2so4? 3) how many moles are there in 2.3 x 1024 atoms of sulfur? 4) how many grams are there in 7.40 x 1023 molecules of ph3? 5) how many moles are there in 7.5 x 1023 molecules of h2so4? 6) how many molecules are there in 55.5 grams of cu(clo¬3)2?

Step-by-step answer

P Answered by Specialist

4

8)2.258×10^24 molecules

Mathematics

100 rules: have to answer correctly have to answer all 16 questions only answer if you only know the answer partial answers or incomplete answers not accepted will be reported point stealers beware 1) how many moles are there in 24.5 grams of cof3? 2) how many molecules are there in 150.00 grams of na2so4? 3) how many moles are there in 2.3 x 1024 atoms of sulfur? 4) how many grams are there in 7.40 x 1023 molecules of ph3? 5) how many moles are there in 7.5 x 1023 molecules of h2so4? 6) how many molecules are there in 55.5 grams of cu(clo¬3)2?

Step-by-step answer

P Answered by Master

4

8)2.258×10^24 molecules

Physics

An iron wire has a cross-sectional area of 5.00 x 10-6 m2. Carry out steps (a) through (e) to compute the drift speed of the conduction electrons in the wire. (a) How many kilograms are there in 1 mole of iron? (b) Starting with the density of iron and the result of part (a), compute the molar density of iron (the number of moles of iron per cubic meter). (c) Calculate the number density of iron atoms using Avogadro’s number. (d) Obtain the number density of conduction electrons given that there are two conduction electrons per iron atom. (e) If

Step-by-step answer

P Answered by Master

Explanation:

Given:

Area, A = 5 × 10^-6 m^2

Current, I = 30 A

U = (m × σ × V)/ρ × e × f × l)

= I/(n × A × Q)

Where,

u is again the drift velocity of the electrons, in m⋅s−1

m is the molecular mass of the metal, in kg

σ is the electric conductivity of the medium at the temperature considered, in S/m.

ΔV is the voltage applied across the conductor, in V

ρ is the density (mass per unit volume) of the conductor, in kg⋅m−3

e is the elementary charge, in C

f is the number of free electrons per atom

ℓ is the length of the conductor, in m

A.

Molar mass of iron = 56 g/mol

Converting g to kg,

56 g/mol × 1 kg/1000 g

= 0.056 kg/mol

= approx. 0.056 kilograms in 1 mole of iron.

B.

Density, ρ = 7874 kg/m^3

Molar density = density, ρ/molar mass

= 7874/0.056

= 1.406 × 10^5 mol/m^3

C.

Avogadros constant, Na = 6.022 × 10^23 atoms/mol

Density of iron atoms = avogadros constant × molar density

= 6.022 × 10^23 × 1.406 × 10^5

= 8.478 × 10^28 atoms/m^3

D.

Fe --> Fe2+ + 2e-

density of conduction electrons = 2 conduction electrons/1 atom of iron

= 2 × 8.478 × 10^28

= 1.69 × 10^29 conduction electrons/m^3

E.

Q = 1.602 × 10^-19 C

Using the equation above,

V = 30/(1.69 × 10^29 × 1.602 × 10^-19 × 5 × 10^-6)

= 2.216 × 10^-4 m/s.

Physics

An iron wire has a cross-sectional area of 5.00 x 10-6 m2. Carry out steps (a) through (e) to compute the drift speed of the conduction electrons in the wire. (a) How many kilograms are there in 1 mole of iron? (b) Starting with the density of iron and the result of part (a), compute the molar density of iron (the number of moles of iron per cubic meter). (c) Calculate the number density of iron atoms using Avogadro’s number. (d) Obtain the number density of conduction electrons given that there are two conduction electrons per iron atom. (e) If

Step-by-step answer

P Answered by Master

Explanation:

Given:

Area, A = 5 × 10^-6 m^2

Current, I = 30 A

U = (m × σ × V)/ρ × e × f × l)

= I/(n × A × Q)

Where,

u is again the drift velocity of the electrons, in m⋅s−1

m is the molecular mass of the metal, in kg

σ is the electric conductivity of the medium at the temperature considered, in S/m.

ΔV is the voltage applied across the conductor, in V

ρ is the density (mass per unit volume) of the conductor, in kg⋅m−3

e is the elementary charge, in C

f is the number of free electrons per atom

ℓ is the length of the conductor, in m

A.

Molar mass of iron = 56 g/mol

Converting g to kg,

56 g/mol × 1 kg/1000 g

= 0.056 kg/mol

= approx. 0.056 kilograms in 1 mole of iron.

B.

Density, ρ = 7874 kg/m^3

Molar density = density, ρ/molar mass

= 7874/0.056

= 1.406 × 10^5 mol/m^3

C.

Avogadros constant, Na = 6.022 × 10^23 atoms/mol

Density of iron atoms = avogadros constant × molar density

= 6.022 × 10^23 × 1.406 × 10^5

= 8.478 × 10^28 atoms/m^3

D.

Fe --> Fe2+ + 2e-

density of conduction electrons = 2 conduction electrons/1 atom of iron

= 2 × 8.478 × 10^28

= 1.69 × 10^29 conduction electrons/m^3

E.

Q = 1.602 × 10^-19 C

Using the equation above,

V = 30/(1.69 × 10^29 × 1.602 × 10^-19 × 5 × 10^-6)

= 2.216 × 10^-4 m/s.

Chemistry

1. How many grams are there in 7.5 x 1023 molecules of H2SO4? 2. How many molecules are there in 122 grams of Cu(NO3)2? 3. How many moles in 28 grams of CO2? 4. What is the mass of 3 moles of Fe2O3? 5. How many moles of neon atoms are present in 16.3 L of neon gas at STP? 6. What is the volume 2 mol of chlorine Cl2 gas at STP 7. Find the mass in grams of 2.00 x 1023 molecules of F2. 8. Determine the volume in liters occupied by 14 g of nitrogen N2 gas at STP. 9. What is the relative density ofnitrogen oxide NO2 gas according to air? 10. What is

Step-by-step answer

P Answered by Master

I cannot give you all the answer but I can help you to solve those.

Explanation:

The first question:

How many grams are there in 7.5× $10^{23}$ molecules of $H_{2} SO_{4}$ ?

So we need to find the molecular mass first, use your periodic table,

And then we can find out 2+32+16×4=98 g/mol

Then, we need to find how many moles, by using Avogadro's constant:

Avogadro's constant: 1 mole = 6.02× $10^{23}$

∴ $\frac{7.5*10^{23} }{6.02*10^{23}}$ =1.25 mol(2d.p.)

Lastly, find the grams using the formula $M= \frac{m}{n}$

m=Mn

m=1.25*98

m=122.5g

-------------------------------------------------------------------------------------------------------------

In conclusion, use those formula to help you:

$M= \frac{m}{n}$ (which M = molecular mass(atomic mass) m=mass of the substance and n = moles)

Avogadro's constant: $\frac {molecular mass} {6.02*10^{23}} = moles$

Chemistry

1. How many grams are there in 7.5 x 1023 molecules of H2SO4? 2. How many molecules are there in 122 grams of Cu(NO3)2? 3. How many moles in 28 grams of CO2? 4. What is the mass of 3 moles of Fe2O3? 5. How many moles of neon atoms are present in 16.3 L of neon gas at STP? 6. What is the volume 2 mol of chlorine Cl2 gas at STP 7. Find the mass in grams of 2.00 x 1023 molecules of F2. 8. Determine the volume in liters occupied by 14 g of nitrogen N2 gas at STP. 9. What is the relative density ofnitrogen oxide NO2 gas according to air? 10. What is

Step-by-step answer

P Answered by Specialist

I cannot give you all the answer but I can help you to solve those.

Explanation:

The first question:

How many grams are there in 7.5× $10^{23}$ molecules of $H_{2} SO_{4}$ ?

So we need to find the molecular mass first, use your periodic table,

And then we can find out 2+32+16×4=98 g/mol

Then, we need to find how many moles, by using Avogadro's constant:

Avogadro's constant: 1 mole = 6.02× $10^{23}$

∴ $\frac{7.5*10^{23} }{6.02*10^{23}}$ =1.25 mol(2d.p.)

Lastly, find the grams using the formula $M= \frac{m}{n}$

m=Mn

m=1.25*98

m=122.5g

-------------------------------------------------------------------------------------------------------------

In conclusion, use those formula to help you:

$M= \frac{m}{n}$ (which M = molecular mass(atomic mass) m=mass of the substance and n = moles)

Avogadro's constant: $\frac {molecular mass} {6.02*10^{23}} = moles$

Chemistry

How many molecules are there in 150g of nano3? 3. how many grams are there in 5.7 * 10^46 molecules of nano3? 4. how many grams are there in 301 miles of water, h2o? 5. how many miles are there in 25g of h2so4? 6. how many molecules are there in 252g of ca(oh)2? 7. how many grams are there in 6.7 * 10^35 atoms of calcium? if you answer this could you put each step too

Step-by-step answer

P Answered by PhD

For the whole set of problems, always remember the Avogadro’s number is 6.023*10^23 units per mole of a substance. Units could be atoms, molecules or formula units.
The first question asks for the number of molecules of NaNO3. The molar mass of NaNO3 is 85 grams per mole. So,
150g NaNO3(1mole NaNO3/85 grams NaNO3)(6.023*10^23 molecules/1mole NaNO3)=1.063*10^24 molecules of NaNO3

5.7*10^46 molecules of NaNO3(1mole NaNO3/6.023*10^23 molecules)(85 grams NaNO3/1mole NaNO3) = 8.044*10^24 grams NaNO3

For the molar mass of water, we have 18.02grams per mole.
301 moles H2O(18.02 grams H2O/1 mole H2O) = 5424.02 grams H2O

For the molar mass of sulfuric acid, we have 98.08 grams per mole.
25g H2SO4(I mole H2SO4/98.08g H2SO4) = 0.2549 mole H2SO4

For the molar mass of Ca(OH)2, we have 74.1 grams per mole.
252gCa(OH)2(1mol/74.1g)(6.023*10^23/1mol) = 2.048*10^24 molecules of Ca(OH)2

For the molar mass of calcium, we have 40 grams of Ca per mole.
6.7*10^35 atoms Ca(1 mole Ca/6.023*10^23 atoms)(40g Ca/1mol Ca) = 4.45*10^13 grams Ca

Chemistry

How many molecules are there in 150g of nano3? 3. how many grams are there in 5.7 * 10^46 molecules of nano3? 4. how many grams are there in 301 miles of water, h2o? 5. how many miles are there in 25g of h2so4? 6. how many molecules are there in 252g of ca(oh)2? 7. how many grams are there in 6.7 * 10^35 atoms of calcium? if you answer this could you put each step too

Step-by-step answer

P Answered by PhD

For the whole set of problems, always remember the Avogadro’s number is 6.023*10^23 units per mole of a substance. Units could be atoms, molecules or formula units.
The first question asks for the number of molecules of NaNO3. The molar mass of NaNO3 is 85 grams per mole. So,
150g NaNO3(1mole NaNO3/85 grams NaNO3)(6.023*10^23 molecules/1mole NaNO3)=1.063*10^24 molecules of NaNO3

5.7*10^46 molecules of NaNO3(1mole NaNO3/6.023*10^23 molecules)(85 grams NaNO3/1mole NaNO3) = 8.044*10^24 grams NaNO3

For the molar mass of water, we have 18.02grams per mole.
301 moles H2O(18.02 grams H2O/1 mole H2O) = 5424.02 grams H2O

For the molar mass of sulfuric acid, we have 98.08 grams per mole.
25g H2SO4(I mole H2SO4/98.08g H2SO4) = 0.2549 mole H2SO4

For the molar mass of Ca(OH)2, we have 74.1 grams per mole.
252gCa(OH)2(1mol/74.1g)(6.023*10^23/1mol) = 2.048*10^24 molecules of Ca(OH)2

For the molar mass of calcium, we have 40 grams of Ca per mole.
6.7*10^35 atoms Ca(1 mole Ca/6.023*10^23 atoms)(40g Ca/1mol Ca) = 4.45*10^13 grams Ca

Chemistry

1how many grams are there in 7.4 x 1023 molecules of agno3? 2 how many grams are there in 7.5 x 1023 molecules of h2so4? 3 how many grams are there in 9.4 x 1025 molecules of h2? plz

Step-by-step answer

P Answered by Specialist

2

In $7,4.10^{23}$ molecules of AgNO_3 are 209 g

In $7,5.10^{23}$ molecules of H_2SO_4 are 123 g

In $9,4.10^{25}$ molecules of H_2 are 312 g

Explanation:

First remember that 1 mole of any element has $6.02 10^{23}$ molecules (this number is a constant).

we know that the molecular mass of an element relates the grams of that substance in 1 mol

To calculate how many grams of AgNO_3 are in $7,4.10^{23}$ molecules we use a rule of three

$6,02.10^{23} molecules \longrightarrow 1mol AgNO_3\\7,4.10^{23} molecules\longrightarrow x\\x=\frac{7.,4.10^{23}molecules.1mol AgNO_3 }{6,02.10^{23}molecules}\\x=1.23 mol AgNO_3$

The molecular mass of AgNO_3 is 169.87 g/mol

$1 mol \longrightarrow 169.87 g\\1.23 mol\longrightarrow x\\x=\frac{1.23 mol. 169.87g}{1 mol} \\x= 209 g AgNO_3$

In $7,4.10^{23}$ molecules of AgNO_3 are 209 g

To calculate how many grams of H_2SO_4 are in $7,5.10^{23}$ molecules we repeat the previous procedure

$6,02.10^{23} molecules \longrightarrow 1mol H_2SO_4\\7,5.10^{23} molecules\longrightarrow x\\x=\frac{7.,5.10^{23}molecules.1mol AgNO_3 }{6,02.10^{23} molecules}\\x=1.25 mol H_2SO_4$

The molecular mass of H_2SO_4 is 98.079 g/mol

$1 mol \longrightarrow 98.079g\\1.25 mol\longrightarrow x\\x=\frac{1.25mol. 98.079g}{1 mol}\\x= 123 g AgNO_3$

In $7,5.10^{23}$ molecules of H_2SO_4 are 123 g

To calculate how many grams of H_2 are in $9,4.10^{25}$ molecules we repeat the previous procedure

$6,02.10^{23} molecules \longrightarrow 1mol H_2\\9,4.10^{25} molecules\longrightarrow x\\x=\frac{9,4.10^{25}molecules.1mol H_2}{6,02.10^{23} molecules}\\x=156 mol H_2$

The molecular mass of H_2 is 2 g/mol

$1 mol \longrightarrow 2g\\156 mol\longrightarrow x \\x=\frac{1.25mol. 98.079g}{1 mol} \\x= 312g$

In $9,4.10^{25}$ molecules of H_2 are 312 g

Chemistry

1how many grams are there in 7.4 x 1023 molecules of agno3? 2 how many grams are there in 7.5 x 1023 molecules of h2so4? 3 how many grams are there in 9.4 x 1025 molecules of h2? plz

Step-by-step answer

P Answered by Master

2

In $7,4.10^{23}$ molecules of AgNO_3 are 209 g

In $7,5.10^{23}$ molecules of H_2SO_4 are 123 g

In $9,4.10^{25}$ molecules of H_2 are 312 g

Explanation:

First remember that 1 mole of any element has $6.02 10^{23}$ molecules (this number is a constant).

we know that the molecular mass of an element relates the grams of that substance in 1 mol

To calculate how many grams of AgNO_3 are in $7,4.10^{23}$ molecules we use a rule of three

$6,02.10^{23} molecules \longrightarrow 1mol AgNO_3\\7,4.10^{23} molecules\longrightarrow x\\x=\frac{7.,4.10^{23}molecules.1mol AgNO_3 }{6,02.10^{23}molecules}\\x=1.23 mol AgNO_3$

The molecular mass of AgNO_3 is 169.87 g/mol

$1 mol \longrightarrow 169.87 g\\1.23 mol\longrightarrow x\\x=\frac{1.23 mol. 169.87g}{1 mol} \\x= 209 g AgNO_3$

In $7,4.10^{23}$ molecules of AgNO_3 are 209 g

To calculate how many grams of H_2SO_4 are in $7,5.10^{23}$ molecules we repeat the previous procedure

$6,02.10^{23} molecules \longrightarrow 1mol H_2SO_4\\7,5.10^{23} molecules\longrightarrow x\\x=\frac{7.,5.10^{23}molecules.1mol AgNO_3 }{6,02.10^{23} molecules}\\x=1.25 mol H_2SO_4$

The molecular mass of H_2SO_4 is 98.079 g/mol

$1 mol \longrightarrow 98.079g\\1.25 mol\longrightarrow x\\x=\frac{1.25mol. 98.079g}{1 mol}\\x= 123 g AgNO_3$

In $7,5.10^{23}$ molecules of H_2SO_4 are 123 g

To calculate how many grams of H_2 are in $9,4.10^{25}$ molecules we repeat the previous procedure

$6,02.10^{23} molecules \longrightarrow 1mol H_2\\9,4.10^{25} molecules\longrightarrow x\\x=\frac{9,4.10^{25}molecules.1mol H_2}{6,02.10^{23} molecules}\\x=156 mol H_2$

The molecular mass of H_2 is 2 g/mol

$1 mol \longrightarrow 2g\\156 mol\longrightarrow x \\x=\frac{1.25mol. 98.079g}{1 mol} \\x= 312g$

In $9,4.10^{25}$ molecules of H_2 are 312 g

How many grams are there in 3 mol NaNO3

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