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Agata [3.3K]
3 years ago
15

Compound A is an organic compound which contains Carbon, Hydrogen and Oxygen. When 0.240g of the vapour of A is slowly passed ov

er a large quantity of heated Copper (II) oxide, CuO, the compound A is completely oxidized to carbon dioxide and water. Copper is the only other product of the reaction. The products are collected and it is found that 0.352g of CO2 and 0.144g of H2O are formed. Questions :1) Calculate the mass of carbon present in 0.352g of CO22) Use this value to calculate the amount in moles of carbon atoms present in 0.240g of A
Chemistry
1 answer:
Keith_Richards [23]3 years ago
7 0

Answer:

1) 0.009 61 g C; 2) 0.008 00 mol C

Step-by-step explanation:

You know that you will need a balanced equation with masses, moles, and molar masses, so gather all the information in one place.

M_r:     12.01               44.01

              C  + ½O₂ ⟶ CO₂

m/g:                            0.352

1) <em>Mass of C </em>

Convert grams of CO₂ to grams of C

44.01 g CO₂ = 12.01 g C

    Mass of C = 0.352 g CO₂ × 12.01 g C/44.01 g CO₂

    Mass of C = 0.009 61 g C

2) <em>Moles of C </em>

Convert mass of C to moles of C.

     1 mol C = 12.01 g C

Moles of C = 0.00961 g C × (1 mol C/12.01 g C)

Moles of C = 0.008 00 mol C

All the carbon comes from Compound A, so there are 0.008 00 mol C in Compound A.

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Reika [66]

Answer: I believe it's C

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3 0
3 years ago
If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr
Alinara [238K]

Answer:

The value of  the heat capacity of the Calorimeter  C_c = 54.4 \frac{J}{c}

Explanation:

Given data

Heat added Q = 4.168 KJ = 4168 J

Mass of water m_w = 75.40 gm

Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c

From the given condition

Q = m_w C_w ΔT + C_c ΔT

Put all the values in above equation we get

4168 = 75.70 × 4.18 × 11.24 +  C_c × 11.24

611.37 =  C_c × 11.24

C_c = 54.4 \frac{J}{c}

This is the value of  the heat capacity of the Calorimeter.

7 0
3 years ago
Br+ is isoelectronic with noble gas?
ArbitrLikvidat [17]

Answer:

yes. I think

Explanation:

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4 0
3 years ago
Gaseous ammonia chemically reacts with oxygen O2 gas to produce nitrogen monoxide gas and water vapor. Calculate the moles of am
mel-nik [20]

Answer:

1.7 moles of ammonia, NH₃.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4NH₃ + 5O₂ —> 4NO + 6H₂O

From the balanced equation above,

4 moles of NH₃ reacted to produce 4 moles of NO.

Finally, we shall determine the number of mole of ammonia, NH₃, needed to produce 1.7 moles of nitrogen monoxide, NO. This can be obtained as follow:

From the balanced equation above,

4 moles of NH₃ reacted to produce 4 moles of NO.

Therefore, 1.7 moles of NH₃ will also react to produce 1.7 moles of NO.

Thus, 1.7 moles of ammonia, NH₃, is required.

5 0
3 years ago
2.50 g of Zn is added to 5.00 g of HCl
Debora [2.8K]
Zn+2HCl ----> 2ZnCl2 + H2

For 2.50 g of Zn

Mass per mol = 2.50/molar mass of Zn = 2.50/65.38 = 0.0382 g/mol
There are two moles of ZnCl2 and total mass = 2*0.0382*molar mass of ZnCl2 = 2*0.0382*136.286 = 10.42 g

For 2 g of HCl

Mass per mol = 2/2*molar mass of HCl = 2/ (2*36.46) = 0.0274 g/mol
For the two moles of ZnCl2, mass produced = 2*0.0274*136.286 = 7.48 g

It can be noted that 2 g of HCl produced less amount of ZnCl and thus it is the limiting reagent.
5 0
3 years ago
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