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Gennadij [26K]
3 years ago
8

The molar solubility of silver bromide, AgBr in pure water is 0.0007350 mol/L. What is the

Chemistry
1 answer:
gayaneshka [121]3 years ago
7 0

Answer:

0.000000540

Explanation:

Step 1: Make an ICE chart for the solution of AgBr

"S" represents the molar solubility of AgBr

        AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)

I                           0             0

C                          +S          +S

E                           S             S

Step 2: Write the expression for the solubility product constant (Ksp)

Ksp = [Ag⁺] [Br⁻] = S × S

Ksp = S² = (0.0007350)² = 0.000000540

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In burning gasoline the products are​
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Answer: petroleum is the main item in it

Explanation:

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3 years ago
Superheated steam at 20 bar and 450oC flows at a rate of 200 kg/min to an adiabatic turbine, where it expands to 10 bar. The tur
sergejj [24]

Explanation:

The given data is as follows.

Pressure of steam at inlet of turbine, P_{1} = 20 bar

Temperature at inlet of turbine, T = 450^{o}C

Pressure at outlet of turbine, P_{2} = 10 bar  

Mass flow rate of steam, m = 200 kg/min

Work produced by the turbine, W_{s} = 1500 kW

Steam is heated at constant pressure to its initial temperature, i.e., temperature at outlet of heat exchanger, T_{3} = 450^{o}C.

(1)   For an adiabatic turbine, the energy balance is  as follows.

           -W_{s} = m({H}_{2} - {H}_{1})

where W_{s} = work done by the turbine

                      m = mass flow rate of steam

H_{1} and H_{2} are the specific enthalpy of steam at inlet and outlet conditions of turbine.

Obtain the specific enthalpy of steam from Properties of Superheated Steam table

At 20 bar and 450^{o}C, H_{1} = 3358 kJ/kg

           H_{2} = H_{1} - \frac{W_{s}}{m}

      H_{2} = 3358 kJ/kg - \frac{1500kJ/s \times 60 s/min}{200 kg/min}

           H_{2} = 2908 kJ/kg

For P = 10 bar, H =2875 kJ/kg for T= 200^{0}C and H = 2975 kJ/kg for T=250^{o}C. Interpolate the values.

The temperature corresponding to P = 10 bar and H_{2} = 2908 kJ/kg is T = 216.5^{o}C

Therefore, the outlet temperature is T_{2} = 216.5^{o}C.

(2)     Energy balance on the heater is  as follows.

         Q = \Delta H = m(H_{3} - H_{2})

where,        Q = heat input required by the steam

                \Delta H = specific enthalpy change

           H_{3} = specific enthalpy of steam at the outlet conditions of heat exchanger

At P = 10 bar and T_{3} = 450^{o}C,  H_{3} = 3371 kJ/kg.

          Q = \frac{200 kg/min}{60 s/min} \times (3371 - 2908)kJ/kg

[/tex]

          Q = 1543.33 kJ/s

or,    Q = 1543.33 kW

Therefore, the heat input required is Q = 1543.33 kW.

8 0
3 years ago
A sample from one of Earth's oceans has a salinity of 34. What is the concentration of dissolved salts in this sample of seawate
11Alexandr11 [23.1K]

Answer:

Concentration of dissolved salts = 34,038.76 ppm

Explanation:

Given:

Salinity of ocean water = 34

Find:

Concentration of dissolved salts

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1g/l = 1001.14 ppm

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3 0
3 years ago
If 1.00 g of a hydrocarbon is combusted and found to produce 3.14 g of co2, what is the empirical formula of the hydrocarbon?
photoshop1234 [79]
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The mass fraction of carbon in CO2 is 3/11. Hence,
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Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g. 

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The empirical formula for the hydrocarbon is therefore, CH₂.
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If you were to take object A to the moon, which has less gravity than Earth, would its mass change? Why or why not?
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No

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<u>Weight</u> changes based on gravity, but mass doesn't

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