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soldier1979 [14.2K]
2 years ago
12

How many atoms of S are in 98.0 grams of hydrosulfuric acid?

Chemistry
1 answer:
noname [10]2 years ago
7 0
<h2>Answer:6.022\times 10^{23}</h2>

Explanation:

The formula for sulphuric acid is H_{2}SO_{4}.

Molecular mass of sulphuric acid is 2(2)+32+16(4)=98g

given weight=98g

Number of moles of sulphuric acid given is \frac{\text{given weight}}{molar mass}=\frac{98}{98}=1

The formula indicates that one mole of H_{2}SO_{4} contains 2 moles of H atoms,1 mole of S atoms,4 moles of O atoms.

So,the compound contains 1 mole of sulphur atoms.

1 mole of sulphur has 6.022\times 10^{23} atoms

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Water flowing at the rate of 13.85 kg/s is to be heated from 54.5 to 87.8°C in a heat exchanger by 54 to 430 kg/h of hot gas flo
weeeeeb [17]

Answer:

=> 572.83 K (299.83°C).

=> 95.86 m^2.

Explanation:

Parameters given are; Water flowing= 13.85 kg/s, temperature of water entering = 54.5°C and the temperature of water going out = 87.8°C, gas flow rate 54,430 kg/h(15.11 kg/s). Temperature of gas coming in = 427°C = 700K, specific heat capacity of hot gas and water = 1.005 kJ/ kg.K and 4.187 KJ/kg. K, overall heat transfer coefficient = Uo = 69.1 W/m^2.K.

Hence;

Mass of hot gas × specific heat capacity of hot gas × change in temperature = mass of water × specific heat capacity of water × change in temperature.

15.11 × 1.005(700K - x ) = 13.85 × 4.187(33.3).

If we solve for x, we will get the value of x to be;

x = 572.83 K (2.99.83°C).

x is the temperature of the exit gas that is 572.83 K(299.83°C).

(b). ∆T = 339.2 - 245.33/ln (339.2/245.33).

∆T = 93.87/ln 1.38.

∆T = 291.521K.

Heat transfer rate= 15.11 × 1.005 × 10^3 (700 - 572.83) = 1931146.394.

heat-transfer area = 1931146.394/69.1 × 291.521.

heat-transfer area= 95.86 m^2.

7 0
2 years ago
How much heat in kJ is produced by the oxidation of 18.6 g of Mn?
soldier1979 [14.2K]
32’ns()3 jsjsusiiaaiaoaoididudud
8 0
2 years ago
Which of the following air pollutants is correctly paired with one of its major effects?
Alik [6]

Answer:

The correct option is;

Sulfur oxides - acid precipitation

Explanation:

Here we have sulfur oxide when present in the atmosphere and mixed with oxygen water and other chemicals form acidic precipitation known as acid rain.

The sulfur oxides reacts with water in the clouds to form sulfuric acid as follows;

The sulfur gas is first oxidized

SO₂ + OH → HOSO₂

The next stage is the formation of sulfur trioxide

HOSO₂ + O₂ → HO₂ + SO₃

Sulfur trioxide combines with water to form sulfuric acid

SO₃ + H₂O → H₂SO₄ (aq).

3 0
2 years ago
Which is an example of ionization? C + O2 mc004-1.jpg CO2 H2CO3 mc004-2.jpg H2O + CO2 SO2 + H2O mc004-3.jpg HSO3– + H+ Mg(OH)2 m
MrMuchimi
Your answer is SO2 + H2O ---> HSOS- + 2OH-

3 0
2 years ago
Read 2 more answers
A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent
Degger [83]

Answer :  The % of (+) limonene isomer = 79%


                The % of (-) limonene isomer = 0%


                The % of enantiomeric excess = 58%


Explanation :   Enantiomeric excess (ee) is the measurement of purity used for chiral substances.


Given,


% of pure limonene enantiomer = The % of (+) limonene isomer = 79%


Therefore, The % of (-) limonene isomer = 0%


Formula used :  

\%(+)\text{ isomer}=\frac{ee}{2}+50\%


Where,         ee → enantiomeric excess


Now, put all the values in above formula, we get the value of enantiomeric excess (ee).


     {ee}=\frac{\%(+)-50\%}{2}


            =\frac{79\%-50\%}{2}


              = 58%



7 0
2 years ago
Read 2 more answers
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