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salantis [7]
2 years ago
5

What is the density of hydrogen sulfide at 1.1 atm and 304K?

Chemistry
1 answer:
Daniel [21]2 years ago
6 0

Use equation:  

PV = nRT  

1.4*V = 1*0.082057*332  

V = 27.24/1.4  

V = 19.46L  

Molar mass H2S = 2+32 = 34g/mol  

34g H2S has volume 19.46L  

Density = mass / volume = 34/19.46 = 1.75 g/L

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What was Thompson’s model of the atom called
shtirl [24]

J. J. Thomson, who discovered the electron in 1897, proposed the plum pudding model of the atom in 1904 before the discovery of the atomic nucleus in order to include the electron in the atomic model. In Thomson's model, the atom is composed of electrons (which Thomson still called “corpuscles,” though G. J.

4 0
3 years ago
Read 2 more answers
2Ag + H2S ? Ag2S + H2 Which statement is true about this chemical equation? A) Ag (silver) is oxidized and H (hydrogen) is reduc
nadezda [96]

Answer:

It is A).

Explanation:

Silver (Ag) goes from the  pure metal to Ag+ losing 1 electron so it is oxidised.

The hydrogen ion  gains electrons and is reduced.

6 0
2 years ago
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Convert the scientific notation to a decimal number. Type the correct answer in the box. 4 × 10-5 cm = cm
MakcuM [25]

Answer:

4 * 10^{-5}cm = 0.00004cm

Explanation:

Given

4 * 10^{-5}cm = [\ ]cm

Required

Solve

Apply law of indices

4 * \frac{1}{10^5}cm = [\ ]cm

Evaluate 10^5

4 * \frac{1}{100000}cm = [\ ]cm

4 * 0.00001cm = [\ ]cm

0.00004cm = [\ ]cm

Hence:

4 * 10^{-5}cm = 0.00004cm

3 0
3 years ago
Two hydraulic cylinders are connected. If the diameter of one piston is twice the other, the how does the pressure experienced b
Aneli [31]

Answer:

They experience the same pressure

Explanation:

To answer this question, we recall Pascal's, Law Pascal's law states that  an increase in pressure at a point in a confined cylinder containing a fluid, there is also an equal increase at all other points in that cylinder.

According to Pascal's law the pressure if the pressure expereienced by the larger diameter piston increases, the pressure experienced by the smaller diameter piston also increases by the same amount

However considering that pressure = Force/area F1/A1 =F2/A2

thus where A1 = πD²÷4 and A2 = πD²÷ 16 we have

we have F1×4/πD² = F2×16/πD² or F1 = 4× F2

They experience the same pressure but the larger cylinder delivers four times the force transmitted from he outside to the smaller cylinder

7 0
3 years ago
A sample of food containing 27 g of fat, 48 g of carbohydrates and 20 g of protein is burned in a bomb calorimeter. In a perfect
rosijanka [135]

Answer:

38.3958 °C  

Explanation:

As,

1 gram of carbohydrates on burning gives 4 kilocalories  of energy

1 gram of protein on burning gives 4 kilocalories  of energy

1 gram of fat on burning gives 9 kilocalories of energy

Thus,

27 g of fat on burning gives 9*27 = 243 kilocalories of energy

20 g of protein on burning gives 4*20 = 80 kilocalories  of energy

48 gram of carbohydrates on burning gives 4*48 = 192 kilocalories  of energy

Total energy = 515 kilocalories

Using,

Q=m_{water}\times C_{water}\times (T_f-T_i)

Given: Volume of water = 23 L = 23×10⁻³ m³

Density=\frac{Mass}{Volume}  

Density of water= 1000 kg/m³

So, mass of the water:  

Mass\ of\ water=Density \times {Volume\ of\ water}  

Mass\ of\ water=1000 kg/m^3 \times {0.023\ m^3}  

Mass of water  = 23 kg

Initial temperature = 16°C  

Specific heat of water = 0.9998 kcal/kg°C  

515=23\times 0.9998\times (T_f-16)

Solving for final temperature as:

<u>Final temperature = 38.3958 °C  </u>

8 0
3 years ago
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