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Leno4ka [110]
2 years ago
11

A 0.15 M NaOH solution has a volume of 0.125 L but is then diluted to 0.15 L. What is the concentration of the new solution

Chemistry
1 answer:
tresset_1 [31]2 years ago
5 0

Answer:

.125 M

Explanation:

.15 M/L   * .125 L = .01875 moles

now dilute to 150 cc  (by adding 25 cc)

.01875M / (150/1000) = .125M

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At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ / mol. When 1.697 g of compound
melisa1 [442]

Answer:

13.85 kJ/°C

-14.89 kJ/g

Explanation:

<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>

<em />

The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

1.697g.\frac{1mol}{101.67g} .\frac{(-3039.0kJ)}{mol} =-50.72kJ

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qcomb + Qcal = 0

Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ

The heat capacity (C) of the calorimeter can be calculated using the following expression.

Qcal = C . ΔT

where,

ΔT is the change in the temperature

Qcal = C . ΔT

50.72 kJ = C . 3.661 °C

C = 13.85 kJ/°C

<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>

Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ

The heat of combustion per gram of B is:

\frac{-56.09 kJ}{3.767g} =-14.89 kJ/g

4 0
3 years ago
Cindy made tea. She started with 300 grams of water at 20 degrees Celsius. She transferred 18,000 calories to the water. What wa
Serjik [45]

Answer:

T final = 80°C

Explanation:

  • Q = mCpΔT

∴ Q = 18000 cal

∴ m H2O = 300 g

∴ Cp H2O (15°C) = 0.99795 cal/g.K ≅ 1 cal/g.K

∴ T1 = 20°C = 293 K

∴ T2 = ?

⇒ 18000 cal = (300 g)(1 cal/g.K)(T2 - 293 K)

⇒ (18000 cal)/(300 cal/K) = T2 - 293 K

⇒ T2 = 293 K + 60 K

⇒ T2 = 353 K (80°C)

8 0
3 years ago
Which is a characteristic of a solution
Annette [7]
 <span>A solution is somthing desolved in somthing else. By desolved i mean it needs to have some particals ionized a solid you place in water that dissosiates (ions split apart from each other) makes a solution a good solution you can make in your kitchen is a salt-water solution, Put some regular table salt in a glass and stir it and you will notice the salt "disapears" what happens is the sodium ions and the chloride Ions seperate and 'hide' between water molocules. 

In basic terms only some substances can make a solutions others are refered to as insoluble as they can't be seperated in water or another solvent. In actuality however all ionic compounds (compounds that are composed of ions) are at least somewhat soluble, but don't dissociate well at all in some solvents. 

Hope that helps</span>
5 0
2 years ago
Please I don't understand any of this. If you can answer by midnight this is due tomorrow, April 29th
kupik [55]

Answer:

(2) Adding more O2(g) would shift the equilibrium to the right because a higher concentration of oxygen is offered than its initial position, therefore more products have to be yielded to maintain equilibrium.

Explanation:

7 0
2 years ago
Various properties of a liquid are recorded during an experiment. the volume of the liquid is measured to be 25.5 milliliters. t
ryzh [129]
The liquid is blue is an example of qualitative data. Qualitative data is data that can not be represented by numbers. The volume and density are both quantitative data.  
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2 years ago
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