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3 years ago

How many moles of MgO can be made from 5 moles of Mg? 2Mg + O2 -> 2MgO

Chemistry

1 answer:

kkurt [141]3 years ago

3 0

Answer:

Explanation:

The expected product is MgO, so the 1-to-1 mole ratio Mg to O in the product is all that is required.

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P4O10(s) = -3110 kJ/mol H2O(l) = -286 kJ/mol H3PO4(s) = -1279 kJ/mol Calculate the change in enthalpy for the following process:

mash [69]

Answer:

-290KJ/mol

Explanation:

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn= 4ΔHH3PO4 - {6ΔHH2O + ΔHP4O10}

ΔHrxn = 4(-1279) - [6(-286) - 3110]

= -5116 -(-1716-3110)

= -5116-(-4826)

= -5116 + 4826 = -290KJ/mol

6 0

3 years ago

Using tongs place the 250 mL beaker on hot plate

LenKa [72]

Answer:

just use the tongs and put it on a plate

Explanation:

4 0

2 years ago

5.22x10-3 = [? ]x 10-2

levacccp [35]

Answer:

5.12

Explanation:

5.22 x 10 - 3 = [?] x 10 - 2 (move - 2 to the other side with changed sign)

5.22 x 10 - 3 + 2 = [?] x 10 (divide both sides by 10)

[?] = (5.22 x 10 - 1) / 10

[?] = 5.12

4 0

3 years ago

Of the groups given, which are the least chemically reactive?

AfilCa [17]

Answer:

Group 8A

Explanation:

This is because the elements in Group 8A is stable, what I mean is this elements have the maximum number of electrons in there last orbit so they dont need to form any compund with any other element in the periodic table.

HOPE THIS WILL HELP YOU❤️

7 0

2 years ago

Please someone help me to solve this problem

Zinaida [17]

Answer:

(a) The final pressure of the sample becomes one-fourth of the original pressure.

(b) The pressure of the sample remains unchanged.

(c) The final pressure of the sample becomes four times of the original pressure.

Explanation:

(a)

$P_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}$

The volume of sample doubled and kelvin temperature halved.

$V_{2}=2V_{1}$

$T_{2}=\frac{1}{2}T_{1}$

$P_{2}=\frac{P_{1}\times V_{1}\times \frac{1}{2}T_{1}}{T_{1}\times2V_{1}}=\frac{P_{1}}{4}$

Therefore, the final pressure of the sample becomes one-fourth of the original pressure.

(b)

Volume and temperature of sample doubled.

$V_{2}=2V_{1}$

$T_{2}=2T_{1}$

$P_{2}=\frac{P_{1}\times V_{1}\times 2T_{1}}{T_{1}\times2V_{1}}={P_{1}}$

Therefore, the pressure of the sample unchanged.

(c)

Volume of sample halved and temperature double.

$V_{2}=\frac{1}{2}V_{1}$

$T_{2}=2T_{1}$

$P_{2}=\frac{P_{1}\times V_{1}\times 2T_{1}}{T_{1}\times \frac{1}{2}V_{1}}={4P_{1}}$

Therefore, the pressure of the sample becomes four times of the original pressure.

6 0

3 years ago