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____ [38]
3 years ago
5

A base has Kb of 2.5 x 10-11. which of the following statements is true?(1 point)

Chemistry
1 answer:
Furkat [3]3 years ago
3 0
I disagree with the answer pick of D. If you have a neutral pH 7 solution and you proceed to add a base even with an relatively insignificantly low Kb your solution would still be more basic then acidic. The answer should be b which is true that the base only ionizes slightly in aqueous solution. This is also truer to the definition of what Kb represents.
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Notice the populations decrease as you go from the bottom of the food chain to the top. Why do you think this is so?
kodGreya [7K]
Exactly what the person above me said
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3 years ago
Read 2 more answers
A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0
3 years ago
Anyone else ever done this? Please help !!
Oksanka [162]

Answers for criss-cross (except Q3) is in the above picture.

5 0
3 years ago
How much nano3 is needed to prepare 225 ml of a 1.55 m solution of nano3?
andrezito [222]

Answer:

= 29.64 g  NaNO3

Explanation:

Molarity is given by the formula;

Molarity = Moles/Volume in liters

Therefore;

Number of moles = Molarity × Volume in liters

                             = 1.55 M × 0.225 L

                             = 0.34875 moles NaNO3

Thus; 0.34875 moles of NaNO3 is needed equivalent to;

   = 0.34875 moles × 84.99 g/mol

   = 29.64 g

5 0
3 years ago
A gas at 42. 0°c occupies a volume of 1. 32 l. if the volume increases to 2. 24 l, what is the new temperature in kelvin?
Nina [5.8K]

the answer is 534.54 kelvin

8 0
2 years ago
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