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BigorU [14]
3 years ago
11

Atoms of which element contain 25 protons?

Chemistry
1 answer:
Law Incorporation [45]3 years ago
4 0
Mangenese is the element which have 25 protons
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What is the mole fraction of NaOH in an aqueous solution that
astraxan [27]

Answer:

The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

Explanation:

We are given that

Aqueous solution that contains 22.9% NaOH by mass means

22.9 g NaOH in 100 g solution.

Mass of NaOH(WB)=22.9 g

Mass of water =100-22.9=77.1

Na=23

O=16

H=1.01

Molar mass of NaOH(MB)=23+16+1.01=40.01

Number of moles =\frac{Given\;mass}{Molar\;mass}

Using the formula

Number of moles of  NaOH(n_B)=\frac{W_B}{M_B}=\frac{22.9}{40.01}

n_B=0.572moles

Molar mass of water=16+2(1.01)=18.02g

Number of moles of water(n_A)=\frac{77.1}{18.02}

n_A=4.279 moles

Now, mole fraction of NaOH

=\frac{n_B}{n_B+n_A}

=\frac{4.279}{0.572+4.279}

=0.882

Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

4 0
3 years ago
How many digits to the right of the decimal point should be used to report the result?
aalyn [17]

Answer:

In engineering and science the common stand is two places.

For example if you get a calculation of 4.567 round up and give the result of 4.57

4 0
3 years ago
Somebody answer please a picture is attached!??
DanielleElmas [232]
9.2 that is the answer
7 0
3 years ago
How many grams of O are in 675 g of Na2O
PIT_PIT [208]
I think it is .00005815
6 0
3 years ago
3.
sineoko [7]

Answer:

3. V = 0.2673 L

4. V = 2.4314 L

5. V = 0.262 L

6. V = 2.224 L

Explanation:

3. assuming ideal gas:

  • PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 225 L

∴ T1 = 175 K

∴ P1 = 150 KPa = 1.48038 atm

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))

⇒ n = 0.043 mol

∴ T2 = 112 K

∴ P2 = P1 = 150 KPa = 1.48038 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)

⇒ V2 = 0.2673 L

4. gas is heated at a constant pressure

∴ T1 = 180 K

∴ P = 1 atm

∴ V1 = 44.8 L

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))

⇒ n = 0.3295 mol

∴ T2 = 90 K

⇒ V2 = RT2n/P

⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)

⇒ V2 = 2.4314 L

5.  V1 = 200 L

∴ P1 = 50 KPa = 0.4935 atm

∴ T1 = 271 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))

⇒ n = 0.2251 mol

∴ P2 = 100 Kpa = 0.9869 atm

∴ T2 = 14 K

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)

⇒ V2 = 0.262 L

6.a)  ∴ V1 = 24.6 L

∴ P1 = 10 atm

∴ T1 = 25°C = 298 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))

⇒ n = 0.0993 mol

∴ T2 = 273 K

∴ P2 = 101.3 KPa = 0.9997 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)

⇒ V2 = 2.224 L

3 0
3 years ago
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