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balandron [24]
2 years ago
11

Summarize the weather conditions related to a cold front. Remember to include all data collected on cold fronts in this activity

to support your answer (examples: interaction of air masses, air pressure, cloud cover, temperature behind/ahead of front, wind direction, precipitation, etc.).
Chemistry
1 answer:
svp [43]2 years ago
5 0

Answer: Many fronts cause weather events such as rain, thunderstorms, gusty winds and  A weather front is a transition zone between two different air masses at the  Lifted warm air ahead of the front produces cumulus or cumulonimbus clouds

Explanation:

Hope this was helpful

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When a 7.00 g7.00 g sample of KBrKBr is dissolved in water in a calorimeter that has a total heat capacity of 2.72 kJ⋅K−1,2.72 k
algol13

Answer : The molar heat of solution of KBr is 19.9 kJ/mol

Explanation :

Mass of KBr = 7.00 g

Molar mass of KBr = 119 g/mole

Heat capacity = 2.72 kJ/K

Change in temperature = 0.430 K

First we have to calculate the moles of KBr.

\text{ Moles of }KBr=\frac{\text{ Mass of }KBr}{\text{ Molar mass of }KBr}=\frac{7.00g}{119g/mole}=0.0588moles

Now we have to calculate the heat of the reaction.

q=c\times \Delta T

where,

q = amount of heat = ?

c = heat capacity = 2.72kJ/K

\Delta T = change in temperature = 0.430 K

Now put all the given values in the above formula, we get:

q=2.72kJ/K\times 0.430K

q=1.17kJ

Now we have to calculate the molar heat of solution of KBr.

\text{Molar heat of solution of }KBr=\frac{q}{n}

where,

n = number of moles of KBr

\text{Molar heat of solution of }KBr=\frac{1.17kJ}{0.0588moles}=19.9kJ/mol

Therefore, the molar heat of solution of KBr is 19.9 kJ/mol

4 0
3 years ago
How many liters of water do we need to add to 5.00 moles to get a 0.648 M solution?
Tanya [424]

Answer: This is a typical acid/base equilibrium problem, that involves the use of logarithms.

Explanation:We assume that both nitric acid and hydrochloric acid dissociate to give stoichiometric

H

3

O

+

.

Moles of nitric acid:

26.0

×

10

−

3

⋅

L

×

8.00

⋅

m

o

l

⋅

L

−

1

=

0.208

⋅

m

o

l

H

N

O

3

(

a

q

)

.

And, moles of hydrochloric acid:

88.0

×

10

−

3

⋅

L

×

5.00

⋅

m

o

l

⋅

L

−

1

=

0.440

⋅

m

o

l

H

C

l

(

a

q

)

.

This molar quantity is diluted to

1.00

L

. Concentration in moles/Litre =

(

0.208

+

0.440

)

⋅

m

o

l

1

L

=

0.648

⋅

m

o

l

⋅

L

−

1

.

Now we know that water undergoes autoprotolysis:

H

2

O

(

l

)

⇌

H

+

+

O

H

−

. This is another equilibrium reaction, and the ion product

[

H

+

]

[

O

H

−

]

=

K

w

. This constant,

K

w

=

10

−

14

at

298

K

.

So

[

H

+

]

=

0.648

⋅

m

o

l

⋅

L

−

1

;

[

O

H

−

]

=

K

w

[

H

+

]

=

10

−

14

0.648

=

?

?

p

H

=

−

log

10

[

H

+

]

=

−

log

10

(

0.648

)

=

?

?

Alternatively, we know further that

p

H

+

p

O

H

=

14

. Once you have

p

H

,

p

O

H

is easy to find. Take the antilogarithm of this to get

[

O

H

−

]

.

Answer link  

4 0
2 years ago
A student has a mixture of solid sand and nacl(aq)in a flask to obtain solid nacl from this mixture the student should
maria [59]
In order to obtain solid NaCl, the student should do a few steps.

First, he/she should do filtration. Pass the mixture through a filter paper, where all the sand should be filtered out already because they're not dissolved in the solution plus they're too small to pass through the filter paper.

Next, the filtrate should be left with NaCl (aqueous state). To seperate NaCl with the liquid, the student can either do evaporation or crystallization, depending on how pure or fast he/she wants the results to be. Evaporation involves heating the beaker or whatever apparatus under the bunsen burner until all the liquid has evaporated. Then, some white powder should be left, they're NaCl solid. For crystallization, the student should just put the beaker on a room condition environment, and wait. They might have to wait a month or so for the liquid to completely evaporate itself and left with clear and pure NaCl crystals.
5 0
3 years ago
How many mm are equal to 21 L. How many mg are equal to 9 g. How many grams are equal to 400 kg
liberstina [14]
<span>How many mm are equal to 21 L? 

1 L = </span>1000000 mm

Convert:-

1000000 × 21 = <span>21000000

21 l = </span><span>21000000 mm

</span><span>How many mg are equal to 9 g?

1 g = 1000 mg

Convert:-

9 </span>× 1000 = 9000

9 g = 9000 mg
8 0
3 years ago
Under what conditions can calcium bromide conduct electricity
Marat540 [252]
CaBr conducts electricity in the molten state but does not conduct as a solid. ionic dissolution equation.
7 0
2 years ago
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