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ANTONII [103]
3 years ago
15

Is 2OH−+Ca2+−>2Ca(OH)2 balanced

Chemistry
2 answers:
wariber [46]3 years ago
3 0

Answer:

No.

Explanation:

No.  There is 1 atom of Ca on the left and 2 Ca's on the right  and 2 OH's on the left and 4 on the right.

The balanced equation is:

4OH-  +  2Ca2+ ---->  2Ca(OH)2.

DENIUS [597]3 years ago
3 0

Answer:

Is not balanced

Explanation:

For an equation to be balanced there must be on each side of the equation the same amount of each atom.

2 OH−  +  Ca 2+   ----> 2 Ca(OH)2

On the left we have the following amounts of atoms:

Ca = 1

O = 2

H = 2

Total charges =  0

And on the right:

Ca = 2

O = 4

H = 4

Total charges =  0

We can see that on the right we have twice as many atoms as on the left, so we can delete a molecule of Ca(OH)2

                         2 OH−  +  Ca 2+   ---->  Ca(OH)2

Now the equation is balanced because there are the same number of atoms on both sides and also the charges are equal in both sides.

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Type the correct answer in the box. Express your answer to three significant figures. This balanced equation shows the reaction
Ratling [72]

Answer:

514.5 g.

Explanation:

  • The balanced equation of the reaction is: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.
  • It is clear that every 2.0 moles of NaOH react with 1.0 mole of H₂SO₄ to produce 1.0 mole of Na₂SO₄ and 2.0 moles of 2H₂O.
  • Since NaOH is in excess, so H₂SO₄  is the limiting reactant.
  • We need to calculate the no. of moles of 355.0 g of H₂SO₄:

n of H₂SO₄ = mass/molar mass = (355.0 g)/(98.0 g/mol) = 3.622 mol.

Using cross multiplication:

∵ 1.0 mol H₂SO₄ produces → 1.0 mol of Na₂SO₄.

∴ 3.622 mol H₂SO₄ produces → 3.662 mol of Na₂SO₄.

  • Now, we can get the theoretical mass of Na₂SO₄:

∴ mass of Na₂SO₄ =  no. of moles x molar mass = (3.662 mol)(142.04 g/mol) = 514.5 g.

8 0
3 years ago
consider the following equilibrium:h2co3(aq) h2o(l) h3o (aq) hco3-1(aq).what is the correct equilibrium expression?
MAVERICK [17]

The equilibrium expression shows the ratio between products and reactants. This expression is equal to the concentration of the products raised to its coefficient divided by the concentration of the reactants raised to its coefficient. The correct equilibrium expression for the given reaction is:<span>

<span>H2CO3(aq) + H2O(l) = H3O+(aq) + HCO3-1(aq)

Kc = [HCO3-1] [H3O+] / [H2O] [H2CO3]</span></span>

4 0
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Identifying map types
Darina [25.2K]

Answer:

Political Map. A political map shows the state and national boundaries of a place. ...

Explanation:

3 0
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Which term represents the fixed proportion of elements in a compound
Gekata [30.6K]

Answer:

Chemical formula.

Explanation:

A chemical formula have fix proportion of atoms of elements.

Chemical formula:

A chemical formula is the way of presenting the chemical proportion of atoms of those elements, that combine to form a compound.

For example:

Water consist of two atom of hydrogen and one atom of oxygen. The atoms of both elements combine to form a compound i.e water. The atoms of both elements always combine in a fixed ratio which is 2:1 in molecule of water.

H₂O

2:1

Carbon dioxide consist of two atom of oxygen and one atom of carbon. The atoms of both elements combine to form a compound i.e carbon dioxide. The atoms of both elements always combine in a fixed ratio which is 1:2 in molecule of carbon dioxide .

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5 0
3 years ago
The chemical formula for magnesium oxide is MgO. A chemist determined by measurements that 0.030 moles of magnesium oxide partic
tamaranim1 [39]

Answer:

1.209g of MgO participates

Explanation:

In this problem, we have 0.030 moles of MgO that participates in a particular reaction.

And we are asked to solve for the mass of MgO that participates, that means, we need to convert moles to grams.

To convert moles to grams we need to use molar mass of the compound:

<em>1 atom of Mg has a molar mass of 24.3g/mol</em>

<em>1 atom of O has a molar mass of 16g/mol</em>

<em />

That means molar mass of MgO is 24.3g/mol + 16g/mol = 40.3g/mol

And mass of 0.030 moles of MgO is:

0.030 moles MgO * (40.3g/mol) =

<h3>1.209g of MgO participates</h3>
3 0
2 years ago
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