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Bumek [7]
3 years ago
11

Measurements show that unknown compound X has the following composition: element mass % carbon 41.0% hydrogen 4.58% oxygen 54.6%

Write the empirical chemical formula of X.
Chemistry
1 answer:
anastassius [24]3 years ago
8 0

Answer:

CHO

Explanation:

Carbon = 41%,  Hydrogen = 4.58%, oxygen = 54.6%

Step 1:

Divide through by their respective relative atomic masses

41/ 12,         4.58/1,         54.6/16

3.41              4.58            3.41

Step 2:

Divide by the lowest ratio:

3.41/3.41,      4.58/3.41,     3.41/3.41

1,                    1,                  1

Hence the empirical formula is CHO

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allochka39001 [22]

Answer:

1^{f} =187.7 \frac{J}{kg}

Explanation:

SO in order to calculate the specific latent heat of fusion, you need to remember the formula:

1^{f} =\frac{Q}{m}

Where 1^{f} representes the specific latent heart of fusion.

Q represents the heat energy added, usually represented in kJ

m represents the mass of the object, in kg.

Now that we have our formula we just have to put our values into the formula:

1^{f} =\frac{Q}{m}

1^{f} =\frac{863kJ}{4.6kg}

1^{f} =187.7 \frac{J}{kg}

SO our answer would be 1^{f} =187.7 \frac{J}{kg}

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3 years ago
Which reactant (X) is missing from the equation shown?<br> X+ PO4 → H3PO4
insens350 [35]

Answer:

H

Explanation:

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3 0
2 years ago
QUESTION 3 (a) Ammonium sulphate, (NH),50, is a soluble salt and it is used in agriculture as fertiliser. 5 g of ammonium sulpha
nataly862011 [7]

Answer:

The equation: (NH₄)₂SO₄ = 2NH4(+) + SO4(-2)

The number of moles = 5 g / 132.14 g/mol = 0.038 mol

The number of molecules = 0.038 X 6.022x10^23 = 2.29x10^23

the number of positive ions present in the ammonium sulphate solution:

2 positive ions for every 1 molecule of (NH₄)₂SO₄

so 2 x 2.29x10^23 = 4.58x10^23

the number of negative ions present in the ammonium sulphate solution

1 negative ion for every 1 molecule of (NH₄)₂SO₄

so 1 x 2.29x10^23 = 2.29x10^23

the total number of ions present in the ammonium sulphate solution​

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4 0
3 years ago
2. A block of aluminum with a mass of 140g is cooled from 98.4oC to 62.2oC with a release of 1137J of heat. From these data, cal
NeTakaya
Q =  M * C *ΔT

Q / <span>ΔT  = M

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<span>
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8 0
3 years ago
At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +
Tju [1.3M]

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(c) A ( g ) + B ( g ) ⟶ C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

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2 years ago
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