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il63 [147K]
2 years ago
12

Determine the number of Mg2+ and PO3−4 ions required to form a neutral ionic compound. Cations and anions

Chemistry
1 answer:
Ierofanga [76]2 years ago
5 0

Answer:

Mg₃(PO₄)₂

Explanation:

Note: The phosphate ion is PO₄³⁻

3Mg⁺² + 2PO₄³⁻ => Mg₃(PO₄)₂

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3 years ago
Epsom salt is commonly purchased in the pharmacy for a variety of uses, anti-inflammatory, laxative, and cosmetic. epsom salt is
Dominik [7]
Epsom salt is MgSO4.
We assume x water of hydration in the crystalline form.

Molecular mass of MgSO4 = 24+32+4*16=120
Molecular mass of MgSO4.xH2O = 120+18x

By proportion, 
2.000/0.977 = (120+18x)/120
Cross multiply
0.977(120+18x) = 120*2.000
from which we solve for x
17.586x+117.24 = 240
x=122.78/17.586
=6.980

Answer: there are 7 water of hydration in Epsom salt, according to the experiment.

Note: more accurate (proper) results may be obtained by using exact values (3-4 significant figures) in the molecular masses.  However, since water of hydration is the nearest integer, using approximate values (to at least two significant figures) suffice.
4 0
3 years ago
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Can anyone please tell me the reaction between...
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hydrogen peroxide on a raw liver bubbles up and stinks.
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4 years ago
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The effectiveness of nitrogen fertilizers depends on both their ability to deliver nitrogen to plants and the amount of nitrogen
densk [106]

Answer:

Ammonia > Urea > Ammonium nitrate > Ammonium sulphate

Explanation:

Percentage by mass of nitrogen in NH3:

Molar mass of NH3= 17 g/mol

Hence % by mass = 14/17 × 100 = 82.35%

% by mass of NH4NO3

Molar mass of NH4NO3 = 80.043 g/mol

Hence; 28/80.043 × 100 = 34.98%

% by mass of (NH4)2SO4;

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% by mass of CH4N2O

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8 0
3 years ago
A 1250.0g block of silver at 1oC is placed in 2.6g of water vapor at 102oC. What is the equilibrium temperature of the two subst
skad [1K]

Answer:

Explanation:

Let final temperature be T .

vapor is at 102⁰C

loss of heat by vapor in turning into water at 100⁰C

= 2.6 x 2 x 1.996 + 2.6 x 2260 = 5886.37 J

loss of heat to lower temperature to T

2.6 x 4.186 x ( 100 - T )

1088.36 - 10.88 T

Total heat loss = 5886.37 + 1088.36 - 10.88 T

= 6974.73 - 10.88 T

heat gain by silver to gain temperature from 1⁰C to T⁰C

= 1250 x ( T - 1 ) x .235 = 293.75 T - 293.75

heat gain = heat loss

293.75 T - 293.75  = 6974.73 - 10.88 T

304.63 T = 7268.48

T = 23.86°C .

6 0
4 years ago
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