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dexar [7]
3 years ago
15

Pls help with science

Chemistry
1 answer:
MissTica3 years ago
6 0
The answer is C because it is believed that those two were once connected
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Indicate if the following statements are TRUE (T) or FALSE (F):
Aleks [24]

Answer:

a) T

b) T

c) F

d) F

e) T

f) T

g) T

h) F

I) F

j) F

k) F

l) F

Explanation:

The w/v concentration is obtained from, mass/volume. Hence;

%w/v= 50/1000= 5%

In the %w/w we have;

25g/100 g = 25% w/w

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Neutralization reaction yields a salt and water

% by mass of carbon is obtained from;

8× 12/114 × 100 = 84.1%

All the ionic substances mentioned have very low solubility in water.

One mole of a substance contains the Avogadro's number of each atom in the compound.

There are two iron atoms so one mole contains 2× 55.85 g of iron.

Some sulphates such as BaSO4 are insoluble in water.

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The equation does not balance with the given coefficients because the number of atoms of each element on both sides differ.

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3 0
3 years ago
Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −
kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)  ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca)  = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207  KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH  of the respective molecules given above,

ΔH  = -1270 + (-393.5) - (-1207)

ΔH  = -456.5 KJ

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