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ruslelena [56]
3 years ago
9

What is the molarity of a NaOH solution if 48.0 mL of a 0.220 M H2SO4 solution is required to neutralize a 25.0-mL sample of the

NaOH solution?
Chemistry
1 answer:
Semenov [28]3 years ago
6 0

Answer:

Molarity NaOH = 0.85M (2 sig figs)

Explanation:

48.0ml(0.220M H₂SO₄) + 25ml(Xmolar NaOH)H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O

2(molarity x volume) H₂SO₄ = (molarity x volume) NaOH

2(0.220M x 48.0ml) = 25.0ml x Molarity NaOH

Molarity NaOH = 2(0.220M x 48.0ml)/25.0ml = 0.8448M ≅ 0.85M (2 sig figs)

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Which water usage uses the least amount of water in a year in the United States
statuscvo [17]

Answer: a

Explanation:

Industry uses only about 18% while the others use around 70-90% of water.

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4 0
2 years ago
1. How many joules must be added to 10.0 g of water to raise its temperature from 10°C to<br> 15°C?
weqwewe [10]

Answer:

209.3 Joules require to raise the temperature from 10 °C to 15 °C.

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m × c × ΔT

Given data:

mass of water = 10 g

initial temperature T1= 10 °C

final temperature T2=  15 °C

temperature change =ΔT= T2-T1 = 15°C - 10°C = 5 °C

Energy or joules added to increase the temperature Q = ?

Solution:

We know that specific heat of water is 4.186 J/g .°C

Q = m × c × ΔT

Q = 10 g × 4.186 J/g .°C × 5 °C

Q = 209.3 J

6 0
3 years ago
What about Kevlar makes it so strong? (1 point)
vfiekz [6]

I believe the answer is (B) as molten Kevlar when spun into fiber, its polymers have a crystalline arrangement.

Hope that helped!

7 0
2 years ago
"what percentage of the solar nebula's mass consisted of elements besides hydrogen and helium gases?"
Alborosie
Your answer would be 2%

6 0
4 years ago
Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

5 0
4 years ago
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