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At constant pressure, which of these systems do work on the surroundings? Check all that apply.

lana66690 [7]

I believe the correct answers from the choices listed above are the second and the last option. At constant pressure, the systems 2A(g) + B(g) ---> 4C(g) and 2C(g) A(s) + B(s) ---> C(g) produces work to the surroundings. <span>When a gas is evolved during a chemical reaction, the gas can be imagined as displacing the atmosphere - pushing it back against the atmospheric pressure. The work done is therefore V*P where V is the volume of gas evolved, and P is the atmospheric pressure. </span>

6 0

3 years ago

you wish to make a 0.161 m nitric acid solution from a stock solution of 3.00 m nitric acid. how much concentrated acid must you

Drupady [299]

From a stock solution of 3.00 m nitric acid, 9.391 ml of stock solution is needed to create a 0.161 m nitric acid solution, which has a total volume of 175 ml of the diluted solution.

A chemical reagent is present in vast quantities as a stock solution. It has a uniform concentration. Examples of typical stock solutions in laboratories are nitric acid and hydrochloric acid. These play a critical role in creating the titration-related solution preparations.

We know the formula for dilution type problems

M1 VI = M2 V 2

Where,

M, = initial molarity

V , = initial Volume

M2 = final molarity

V 2 = final Volume

Hene given -

M, = 3.00 M

VI = ?

M2 = 0.161M

V 2 = 175 ml

Accordingly ' MI V1 = M2 V 2

V1 = $(M2*V2)/M1$

V1= (0.161M*175ml)/ 3.00M

v1 = 9.391

The required volume of Stock solution is 9.391ml.

Learn more about Stock solution here

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7 0

1 year ago

A chemical compound has a molecular weight of 89.05 g/mole. 1.400 grams of this compound underwent complete combustion under con

Nataly_w [17]

Answer:

$\Delta _{comb}H=-2,265\frac{kJ}{mol}$

Explanation:

Hello!

In this case, for such calorimetry problem, we can notice that the combustion of the compound releases the heat which causes the increase of the temperature by 11.95 °C, it means that we can write:

$Q _{comb}=-C_{calorimeter}\Delta T_{calorimeter}$

In such a way, we can compute the total released heat due to the combustion considering the calorimeter specific heat and the temperature raise:

$Q _{comb}=-2980\frac{J}{\°C} *11.95\°C\\\\Q _{comb}=-35,611J$

Next, we compute the molar heat of combustion of the compound by dividing by the moles, considering 1.400 g were combusted:

$n=1.400g*\frac{1mol}{89.05g} =0.01572mol$

Thus, we obtain:

$\Delta _{comb}H=\frac{Q_{comb}}{n}=\frac{-35,611J}{0.01572mol} \\\\\Delta _{comb}H=-2,265,331\frac{J}{mol}*\frac{1kJ}{1000J} \\\\\Delta _{comb}H=-2,265\frac{kJ}{mol}$