CAN SOMEONE PLEASE WRITE ME SHORT STORY FOR THIS. I'LL GIVE YOU BRAINLIEST. I PROMISE. 98 POINTS

What does the atomic number of an element refer to on the periodic table?

jekas [21]

The elements are listed in order of increasing the atomic number. Its the number that is the number of protons in the nucleus of an atom.

6 0

3 years ago

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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

1 year ago

How many visible impact craters does the moon contain?

frez [133]

I believe the answer is B. a million or more.

5 0

3 years ago

Which table correctly identifies the subatomic particle's charge and mass? a b c d

s344n2d4d5 [400]

It’s b that charts say it

8 0

3 years ago

The substance water always has a mass ratio of 11% H to 89% O. If 5.00g of a substance containing H and O was decomposed into .2

Gre4nikov [31]

Answer:

No the substance is not water.

Explanation:

The balance chemical equation for the decomposition of water is as follow;

2 H₂O = 2 H₂ + O₂

Step 1: Calculate moles of H₂O;

Moles = Mass / M.Mass

Moles = 5.0 g / 18.01 g/mol

Moles = 0.277 moles of H₂O

Step 2: Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:

According to equation,

2 moles of H₂O produced = 1 mole of O₂

So,

0.277 moles of H₂O will produce = X moles of O₂

Solving for X,

X = 0.277 mol × 1 mol / 2 mol

X = 0.138 moles of O₂

Also,

According to equation,

2 moles of H₂O produced = 2 mole of H₂

So,

0.277 moles of H₂O will produce = X moles of H₂

Solving for X,

X = 0.277 mol × 2 mol / 2 mol

X = 0.227 moles of H₂

Step 3: Calculate Mass of O₂ and H₂ as;

For O₂:

Mass = Moles × M.Mass

Mass = 0.138 mol × 31.99 g/mol

Mass = 4.44 g of O₂

For H₂:

Mass = Moles × M.Mass

Mass = 0.227 mol × 2.01 g/mol

Mass = 0.559 g of H₂

Conclusion:

From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.

6 0

3 years ago