3 years ago

How much heat is absorbed when 15.0 g of liquid water is

Chemistry

1 answer:

NeX [460]3 years ago

8 0

Answer:

376.56

Explanation:

using the formula q=mcΔT, where m=15g, C=4.18J/g°C and ΔT=6.0°C

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How many moles of electrons is required to deposit 5.6g of iron from a solution of iron (2) tetraoxosulphate(6)

yuradex [85]

Answer:

0.20 mol

Explanation:

Let's consider the reduction of iron from an aqueous solution of iron (II).

Fe²⁺ + 2 e⁻ ⇒ Fe

The molar mass of Fe is 55.85 g/mol. The moles corresponding to 5.6 g of Fe are:

5.6 g × 1 mol/55.85 g = 0.10 mol

2 moles of electrons are required to deposit 1 mole of Fe. The moles of electrons required to deposit 0.10 moles of Fe are

0.10 mol Fe × 2 mol e⁻/1 mol Fe = 0.20 mol e⁻

7 0

3 years ago

I will give brainliest!!! In the reaction represented by the equation Al2O3 -> Al + O2, what is the mole ratio of aluminum to

Effectus [21]

Answer:;

4:3

Explanation:

Balanced:

4Al + 302 yields 2al2O3

7 0

3 years ago

Which of the following are potential benefits of wind turbines? (may be more than one answer)

Inga [223]

All except for C. And it's good for the environment.

7 0

3 years ago

Under the right conditions aluminum will react with chlorine to produce aluminum chloride.

salantis [7]

Answer:

$m_{Al}=9.42gAl$

Explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

$m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2} *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl$

Regards!

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3 years ago

(First to answer gets some good) Order the following components of the universe from largest to smallest: Moon, planet,universe,

cluponka [151]

Answer:

Universe, galaxy, solar system, star, planet, moon and asteroid.

Explanation:

You're welcome!

7 0

3 years ago