PLEASE HELP QUIKLY!

Suppose 50.0g of silver nitrate is reacted with 50g of hydrochloric acid producing silver chloride and a mixture of other produc

docker41 [41]

Answer:

53.6 grams of silver chloride was produced.

Explanation:

$AgNO_3+HCl+\rightarrow AgCl+HNO_3$

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

Mass of silver nitrate = 50.0 g

Mass of hydrogen chloride = 50.0 g

Mass of silver chloride = x

Mass of nitric acid = 46.4 g

Mass of silver nitrate + Mass of hydrogen chloride =

Mass of silver chloride + Mass of nitric acid

[te]50.0 g+50.0 g=x+46.4 g[/tex]

$x=50.0 g+50.0 g - 46.4 g = 53.6 g$

53.6 grams of silver chloride was produced.

8 0

3 years ago

Hydroelectric power plants capture the ______________ of falling water to generate electricity. The falling water turns a wheel

LiRa [457]

Answer:

1: energy

2: converts

Kinetic energy is either 3 or 4, those questions are really similar.

3 0

2 years ago

A sample of water at 100°C is converted to steam after absorbing 820 kJ of heat. How grams of H2O are contained in the sample? A

Varvara68 [4.7K]

Since water is already at 100°C all the energy is used to evaporate it.
Now we can calculate how many mols of water are evaporated with 820kJ.
$N= \frac{820}{41} =20 mol$
We calculated that we got 20 mols of water evaporated. Now, all we have to do is find how many grams is a mol of water. Molar mass of water is 20.16 g/mol.
The final answer is:
$m=20*20.16=403.2g$

7 0

3 years ago

What are the first two quantum numbers for the electrons located in subshell 4d?

SCORPION-xisa [38]

Answer:

n = 4, l = 2

Explanation:

The number 4 in 4d is the principal quantum number (n).

The letter d in 4d tells us that we have a d orbital, as determined by the secondary quantum number (l).

The quantum number l tells us the shape of the orbital.

l = 0 s orbital

l = 1 p orbital

l = 2 d orbital

7 0

3 years ago

Read 2 more answers

How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles

sattari [20]

1) Balance the chemical equation.

$3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2$

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4$

We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. This is the excess reactant.

3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2$

We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

$mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2$

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

$g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}$ $g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2$

The mass of Ca3(PO4)2 produced is 351 g Ca3(PO4)2.

Option D.

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8 0

11 months ago