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Flura [38]
3 years ago
13

Can someone help me with chemistry?

Chemistry
2 answers:
mihalych1998 [28]3 years ago
8 0

<span>2AlPO4 ( aq) + 3MgCl2 (aq) -> Mg3(PO4)2 (s) + 2AlCl3 (aq) </span>

<span>Right answer is D
 </span>

Makovka662 [10]3 years ago
5 0

Answer : Yes, the given reaction is the complete balanced equation.

Explanation :

Balanced equation : It is defined as the number of individual atoms of an element present on the reactant side always be equal to the number of individual atoms of an element present on product side. That means the complete balanced equation are those which follows the law of conservation of mass.

When aluminum phosphate reacts with magnesium chloride in an aqueous solution then it gives magnesium phosphate and aluminium chloride as a product.

The complete balanced equation will be,

2AlPO_4(aq)+3MgCl_2(aq)\rightarrow Mg_3(PO_4)_2(s)+2AlCl_3(aq)

By the stoichiometry, 2 moles of aluminium phosphate react with 3 moles of magnesium chloride to give 1 mole of magnesium phosphate and 2 moles of aluminium chloride.

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a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re
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Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

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moles of H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61

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Now put all the given values in the formula of molality, we get

Molality=\frac{0.61\times 1000}{54.6ml}=11.2M

To calculate the volume of acid, we use the equation given by neutralisation reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock acid which is H_2SO_4

M_2\text{ and }V_2 are the molarity and volume of dilute acid which is H_2SO_4

We are given:

M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL

Putting values in above equation, we get:

11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

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