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kakasveta [241]
3 years ago
13

Cacl2+na3po4 ca3po42 + nacl

Chemistry
1 answer:
Alex787 [66]3 years ago
5 0
<h3>Answer:</h3>

3CaCl₂ + 2Na₃PO₄→ Ca₃(PO₄)₂ + 6NaCl

<h3>Explanation:</h3>

We are given the Equation;

CaCl₂ + Na₃PO₄→ Ca₃(PO₄)₂ + NaCl

Assuming the question requires us to balance the equation;

  • A balanced chemical equation is one that has equal number of atoms of each element on both sides of the equation.
  • Balancing chemical equations ensures that they obey the law of conservation of mass in chemical equations.
  • According to the law of conservation of mass in chemical equation, the mass of the reactants should always be equal to the mass of the products.
  • Balancing chemical equations involves putting appropriate coefficients on the reactants and products.

In this case;

  • To balance the equation we are going to put the coefficients 3, 2, 1, and 6.
  • Therefore; the balanced equation will be;

3CaCl₂ + 2Na₃PO₄→ Ca₃(PO₄)₂ + 6NaCl

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Under which conditions will sugar most likely dissolve fastest in a cup of water?
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If the cup of water contained hot water. Sugar will dissolve a lot slower if it was in room temperature and even slower if it was in cold water.
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A train in Japan can travel 813.5 miles in 5 hours
Anastaziya [24]

Answer:

162.7miles/hr

Explanation:

Given parameters:

Distance covered by the train  = 813.5miles

Time taken  = 5hours

Unknown:

Speed of the train  = ?

Solution:

Speed is a physical quantity.

It is mathematically expressed as;

      Speed  = \frac{distance}{time}

So, input parameters and solve;

   Speed  = \frac{813.5}{5}  = 162.7miles/hr

8 0
3 years ago
How many grams are in 3.21 x 1024 molecules of potassium hydroxide?
MakcuM [25]

Answer:

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Explanation:

3 0
3 years ago
A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its
Over [174]
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

4 0
3 years ago
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