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kirza4 [7]
3 years ago
11

Molecule contains carbon, hydrogen and sulfur atoms. When a sample of 0.535g of this compound is burnt in oxygen, 1.119 g of CO2

and 0.229 gof H2O and 0.407g of SO2are obtained.
Calculate its empirical formula.
Chemistry
1 answer:
OLga [1]3 years ago
4 0

Answer:

The empirical formula is, C4H4S

Explanation:

Number of moles of carbon = 1.119 g/ 44g/mol = 0.025 moles

Mass of Carbon= 0.025 moles × 12 g/ mole = 0.3 g

Number of moles of hydrogen = 0.229/18g/mol × 2 = 0.025 moles

Mass of hydrogen = 0.025 moles × 1 = 0.025 g

Number of moles of sulphur = 0.407g/ 64 g/mol = 0.0064 moles

Mass of sulphur= 0.0064 moles ×32 = 0.2 g

Now we obtain the mole ratios by dividing through by the lowest ratio.

C- 0.025 moles/ 0.0064 moles, H- 0.025 moles/ 0.0064 moles, S- 0.0064 moles/0.0064 moles

C4H4S

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Answer:

b i did the test

Explanation:

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3 years ago
Rank the elements according to highest ionization energy:<br><br> Be, C, O, Ne, B, Li, F, N
aleksklad [387]
Given:
Be - Beryllium   -   9,3227
C   - Carbon      - 11,2603
O   - Oxygen      - 13,6181
Ne - Neon          - 21,5645
B   - Boron          -   8,298
Li  - Lithium        -   5,3917
F   - Fluorine      - 17,4228
N   - Nitrogen    - 14,5341

Arranged from highest ionization energy to lowest ionization energy.

Ne ; F ; N ; O ; C ; Be ; B ; Li
5 0
3 years ago
Read 2 more answers
A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0
bija089 [108]

Answer:

E = 0.062 V

Explanation:

(a) See the attached file for the answer

(b)

Calculating the voltage (E) using the formula;

E = - (2.303RT/nf)log Cathode/Anode

Where,

R = 8.314 J/K/mol

T = 35°C = 308 K

F- Faraday's constant = 96500 C/mol,

n = number of moles of electron = 2

Substituting, we have

E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)

   = -0.031 * -2

  = 0.062V

Therefore, the voltmeter will show a voltage of 0.062 V

5 0
3 years ago
A sample of pure calcium fluoride with a mass of 15.0 g contains 7.70 g of calcium. how much calcium is contained in 45.0 g of c
kolbaska11 [484]

In a sample of pure calcium fluoride of mass 15.0 g, 7.70 g of calcium is present. First convert the mass into number of moles as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass.

Molar mass of Ca is 40 g/mol, putting the values,

n=\frac{7.70 g}{40 g/mol}=0.1925 mol

Similarly, molar mass of CaF_{2} is 78.07 g/mol thus, number of moles will be:

n=\frac{15.0 g}{78.07 g/mol}=0.1921 mol.

Thus, 0.1921 mol of CaF_{2} have 0.1925 mol of Ca, or 1 mole of CaF_{2} will have approximately 1 mole of Ca.

Now, mass of Ca needs to be calculated in 45.0 g of CaF_{2}. Converting mass into number of moles first,

n=\frac{45.0 g}{78.07 g/mol}=0.5764 mol

Thus, number of moles of Ca will also be 0.5764 mol, converting number of moles into mass,

m=n\times M=0.5764 mol\times 40 g/mol=23.06 g

Therefore, mass of Ca will be 23.06 g.

6 0
3 years ago
What are two factors that affect the function of enzymes
lapo4ka [179]

Answer:

Temperature

PH

Explanation:

Enzymes are organic catalysts that speeds up the rate of a chemical reaction. Their activities are affected by temperature and pH of the environment among other factors.

  • At extreme temperatures, enzymes can easily be denatured and they will cease to act.
  • At extremes of pH, they can also be adversely affect.

Enzymes generally have temperature and pH window for their activity.

4 0
3 years ago
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