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3 years ago

What is the silver concentration in a solution prepared by mixing 250

Chemistry

1 answer:

kotegsom [21]3 years ago

6 0

Answer:

0.221M

Explanation:

From the question ,

The Molarity of AgNO₂ = 0.310 M

Hence , the concentration of Ag⁺ = 0.301 M

The volume of AgNO₂ = 250 mL

and,

The Molarity of Sodium chromate = 0.160 M

The volume of Sodium chromate = 100 mL.

As the solution is mixed the final volume becomes ,

250mL +100mL = 350mL

Now, using the formula , to find the final molarity of the mixture ,

M₁V₁ ( initial ) = M₂V₂ ( final )

substituting the values , in the above equation ,

0.310M * 250ml = M₂ * 350ml

M₂ = 0.221M

Hence , the concentration of the silver in the final solution = 0.221M

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Ghella [55]

Answer: The enthalpy change is 34.3 kJ

Explanation:

The conversions involved in this process are :

$(1):H_2O(s)(-18^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(25^0C)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change = ?

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$c_{l}$ = specific heat of liquid water = $4.184J/g^0C$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6010 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]$ $\Delta H=34279.8J=34.3kJ$ (1 KJ = 1000 J)

Therefore, the enthalpy change is 34.3 kJ

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a car is traveling at a speed of 55 mph the speed increased to 75 mph if the time needed to hange its speed is 4 seconds whats t

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3 years ago

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