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Nimfa-mama [501]
3 years ago
13

H3PO4 + Ca(OH)2 → Ca(H2PO4)2 + H2O

Chemistry
1 answer:
aniked [119]3 years ago
8 0

Given question is incomplete. The complete question is as follows.

Balance the following equation:

H_3PO_4 + Ca(OH)_2 \rightarrow Ca(H_2PO_4)_2 + H_2O

Answer: The balanced chemical equation is as follows.

2H_3PO_4 + Ca(OH)_2 \rightarrow Ca(H_2PO_4)_2 + 2H_2O

Explanation:

When a chemical equation contains same number of atoms on both reactant and product side then this equation is known as balanced equation.

For example, H_3PO_4 + Ca(OH)_2 \rightarrow Ca(H_2PO_4)_2 + H_2O

Number of atoms on reactant side:

H = 5

P = 1

O = 6

Ca = 1

Number of atoms on product side:

H = 6

P = 2

O = 9

Ca = 1

In order to balance this equation, we will multiply H_3PO_4 by 2 on reactant side and we will multiply H_2O by 2 on product side. Hence, the balanced chemical equation is as follows.

2H_3PO_4 + Ca(OH)_2 \rightarrow Ca(H_2PO_4)_2 + 2H_2O

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N2-3H2 → 2NH3
aniked [119]

Answer:

6 mols H2 are needed

Explanation:

N2 = 28.01g/mol

H2 = 2.02g/mol

\frac{2 mol N_{2} }{1} * \frac{3 mol H_{2}  }{ 1 mol N_{2} } = 6 mol H2

8 0
3 years ago
Bilangan oksidasi vanadium paling tinggi terdapat dalam senyawa..
emmasim [6.3K]
The answer is (b). As, vanadium is attached to five fluoride atoms, each flouride containing -1 oxidation state, hence five fluoride contains -5, to neutralize, vanadium should have +5 oxidation state.

8 0
3 years ago
What is the mass of insoluble lead(II) iodide (461.0 g/mol) produced from 0.830 g of potassium iodide (166.00 g/mol) and aqueous
VikaD [51]

Answer:

C

Explanation:

7 0
3 years ago
When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.
Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2

And the resulting percent yield:

Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%

Regards!

4 0
3 years ago
5) Some students believe that teachers are full of hot air. If I inhale 2.2 liters of gas at a temperature of 180 C and it heats
ahrayia [7]

Given :

If I inhale 2.2 liters of gas at a temperature of 180 C and it heats to a temperature of 380 C in my lungs.

To Find :

The new volume of the gas.

Solution :

Since, their is no information about pressure, so we will assume that pressure is constant.

We know, relation between temperature and volume in constant pressure is :

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

Putting all given values in above equation, we get :

V_2 = \dfrac{V_1\times T_2}{T_1}\\\\V_2 = \dfrac{2.2\times (180+273)}{(380+273)}\\\\V_2 = 1.53 \ liters

Hence, this is the required solution.

8 0
3 years ago
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