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rodikova [14]
2 years ago
8

1. What causes frost to form on the outside of acold container.​

Chemistry
1 answer:
Kamila [148]2 years ago
6 0

Answer:

condensation

Explanation:

Frost forms when an outside surface cools past the dew point. The dew point is the point where the air gets so cold, the water vapor in the atmosphere turns into liquid. This liquid freezes. If it gets cold enough, little bits of ice, or frost, form

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What ion is at a higher concentration in the extracellular fluid than in the intracellular fluid?
Phoenix [80]
There are two types of fluid in the body extracellular fluid and intracellular fluid (ECF and ICF), together they are account for total body water.
The Sodium (Na+) ion is at higher concentration in the extracellular fluid than in the intracellular fluid. The function of extracellular fluid is that it provide cells to watery environment so that they can easily live and perform their function.
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3 years ago
Drag the item from the idea bank to its corresponding match
babymother [125]
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5 0
3 years ago
PLEASE HELP 50 POINTS A really colorful reaction is the synthesis reaction of aluminum and iodine, often called the “Purple Haze
marysya [2.9K]

Answer:

I know that the limited reactant Al.

7 0
3 years ago
How many milliliters of 4.00 M NaOH are required to exactly neutralize 50.0 milliliters of a 2.00 M solution of HNO3 ?
kramer

Answer: The volume of NaOH required is 25.0 ml

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity HNO_3 = 1

M_1 = molarity of HNO_3 solution = 2.00 M

V_1 = volume of  HNO_3 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 4.00 M

V_1 = volume of  NaOH solution =  ?

Putting in the values we get:

1\times 2.00\times 50.0=1\times 4.00\times V_2

V_2=25.0ml

Therefore, volume of NaOH required is 25.0 ml

3 0
3 years ago
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Lera25 [3.4K]

Answer:

NH3

Explanation:

2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)

So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.

n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol

n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol

From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2

4 0
3 years ago
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