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zvonat [6]
3 years ago
13

What is the correct ionic formula when mg2+ and p3− react?

Chemistry
2 answers:
krek1111 [17]3 years ago
8 0
Any compound should have a neutral charge. So the compound you make should have a net charge of zero.

So you have:
Mg which is 2+
And P which is 3-

You want to balance out both their charges. If you simply made MgP it would have a negative charge (because 2-3 = -1.)

Now to balance this, you have to find the LCM: In this case it is 6. (2x3)

Mg: Mg x 3
P: P × 2

You'll finally get Mg3P2.
Mumz [18]3 years ago
6 0

Answer: Ionic formula will be Mg_3P_2.

Explanation: Mg^{2+} and P^{3-} ions will form a ionic compound. Ionic compounds have both metals and non-metals.

Here Mg^{2+} is a metal and P^{3-} is a non-metal.

The net charge on any compound must be 0.

So we need 2 phosphate ions to balance the charge on Mg^{2+} ions. Similarly we need 3 Magnesium ions to balance the charge on P^{3-} ions.

Criss-crossing the charges, we will get the formula as Mg_3P_2

Criss-crossing is shown in the image below.

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fgiga [73]
150/30 = 5
HF1 20/2 = 10
HF2 10/2 = 5
HF3 5/2 = 2.5
HF4 2.5/2 = 1.25
HF5 1.25/2 = 0.625
Answer: 0.63g
5 0
2 years ago
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Use the formation reactions below such that when added together, they match the balanced equation for the combustion of methane.
muminat

Answer:

ΔH of the reaction is -802.3kJ.

Explanation:

Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.

Using the reactions:

<em>(1) </em>Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ

<em>(2) </em>Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k J

<em>(3) </em>H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ

The sum of (2) - (1) produce:

CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ

And the sum of this reaction with 2×(3) produce:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =

<em>-802.3kJ</em>

7 0
3 years ago
Calculate number of moles in 10.6g of sodium carbonate
natulia [17]
<em>m Na₂CO₃: 23g×2 + 12g + 16g×3 = 106 g/mol</em>
------------------------------
1 mol ------- 106g 
X ------------ 10,6g
X = 10,6/106
<u>X = 0,1 mol Na₂CO₃</u>
7 0
3 years ago
2. A quantity of 1.922g of methanol (CH3OH) was burned in a constant-volume
Cerrena [4.2K]
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
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3 years ago
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Ivanshal [37]

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