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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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4 0
3 years ago
(a) carbonate buffers are important in regulating the ph of blood at 7.40. what is the concentration ratio of co2 (usually writt
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A) when the balanced equation of the reaction is:
H2CO3(aq) → HCO3 -(aq) + (H+)
and when we have Ka = 4.3 x10^-7 & PH = 7.4 
So first we will get PKa = -㏒ Ka
PKa = -㏒(4.3x10^-7) = 6.37 by substitution with Pka value in the following formula:
PH = Pka + ㏒[salt/acid]
PH= PKa + ㏒[HCO3-]/[H2CO3]
㏒[HCO3-]/[H2CO3] = PH-Pka
[HCO3-] /[H2CO3] = 10^(7.4 - 6.37)
∴[HCO3-]/[H2CO3] = 11.7
∴[H2CO3]/[HCO3-] = 1/11.7 =  0.09

B) when The balanced equation for this reaction is:
H2PO42-(aq) → HPO4-(aq) + H+
and when we have Ka = 6.2x10^-8 & PH = 7.15
So Pka= -㏒Ka = -㏒(6.2x10^-8) = 7.2 by substitution by Pka value in the following formula:
PH = Pka + ㏒[salt/acid]
7.15= 7.2 + ㏒[HPO4]/[H2PO4] 
-0.05 = ㏒[HPO4]/[H2PO4]
∴[HPO4]/[H2PO4] = 10^-0.05 = 0.89
∴[H2PO4]/[HPO4] = 1/0.89 = 1.12

c) H3PO4(aq) ↔ H2PO-(aq) + H+
the answer is: because we have Ka =7.5x10^-3 and it is a high value of Ka to make a good buffer, also we need a week acid with th salt of the week acid as H3PO4 is a strong acid so it does'nt make a goof buffer.


6 0
3 years ago
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