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3 years ago

13

Fill in the blanks: compounds break up into and in solution, which can move freely and carry an _ cur

rent.

1 answer:

zmey [24]3 years ago

7 0

I hate virtual school

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A student determines the iron(III) content of a solution by first precipitating it as iron(III) hydroxide, and then decomposing

Semmy [17]

Answer:

3.3535 g

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For iron(III) nitrate :

Molarity = 0.404 M

Volume = 52.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 52.0×10⁻³ L

Thus, moles of iron(III) nitrate :

$Moles=0.404 \times {52.0\times 10^{-3}}\ moles$

Moles of iron(III) nitrate = 0.021 moles

1 mole of iron(III) nitrate forms 1 mole of iron(III) hydroxide which further forms 1 mole of iron(III) oxide.

Thus, moles of iron(III) oxide formed from 0.021 moles of iron(III) nitrate = 0.021 moles

Also, molar mass of iron(III) oxide = 159.69 g/mol

So, mass of iron(III) oxide = 0.021 moles × 159.69 g/mol = 3.3535 g

6 0

2 years ago

Which are functions of the stomata in the leaves of seed plants? Check all that apply.

viktelen [127]

Answer:

Take in carbon dioxide, reduces water loss, and produce oxygen.

Explanation: Hi! I've learned about these things in the past, really tough stuff! Hopefully this helps, I believe the three options above are correct, only one I'm possibly questioning is producing oxygen. Good luck!

3 0

3 years ago

Read 2 more answers

How can parents and other caregivers help children learn good nutrition in the home?

ValentinkaMS [17]

Answer:

Healthy Eating — Make sure your child's caregivers offer healthy snacks and meals. Consider other influences. Just as they do for you, your children's friends and the media can also affect healthy choices. Some TV, online, and other ads try to persuade children to consume high-fat foods and sugary drinks.

4 0

3 years ago

Calculate the heat needed to increase the temperature of 100. g water from 45.7 C to 103.5 C.

MA_775_DIABLO [31]

Answer:

Total heat required to raise the temperature of water from 45.7°C to 103.5°C

= 249,362.4 J

Explanation:

The Heat required to raise the temperature of 100.0 g of water from 45.7°C to 103.5°C will be a sum of;

- The heat required to raise the 100 g of water from 45.7°C to water's boiling point of 100°C

- The Heat required to vaporize the 100 g of water at its boiling point

- The Heat required to raise the temperature of this vapour from 100°C to 103.5°C

1) The heat required to raise the 100 g of water from 45.7°C to water's boiling point of 100°C

Q = mCΔT

m = 100 g

C = 4.18 J/g.°C

ΔT = change in temperature = (100 - 45.7) = 54.3°C

Q = 100 × 4.18 × 54.3 = 22,697.4 J

2) The Heat required to vaporize the 100 g of water at its boiling point

Q = mL

m = 100 g

L = ΔHvaporization = 2260 J/g

Q = mL = 100 × 2260 = 226,000 J

3) The Heat required to raise the temperature of this vapour from 100°C to 103.5°C

Q = mCΔT

m = 100 g

C = 1.90 J/g.°C

ΔT = change in temperature = (103.5 - 100) = 3.5°C

Q = 100 × 1.9 × 3.5 = 665 J

Total heat required to raise the temperature of water from 45.7°C to 103.5°C

= 22,697.4 + 226,000 + 665

= 249,362.4 J

Hope this Helps!!!

4 0

3 years ago

Read 2 more answers

ATOMS WE LOVE ATOMS, answer this and get points , also thx for your help :)

LuckyWell [14K]

Answer:C because they have to I have the same mass before and after the equation.

Explanation:

HOPE THIS HELPED!! :) ;) <3<3

7 0

2 years ago

Read 2 more answers