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4 years ago

Water condenses onto grass in the early morning when the air is clear because...

2 answers:

8 0

The answer is b i know what courses you are taking and i can remember it clearly saying in the morning when there is small droplets on the grass and leaves on the trees it is a result of the air close to the ground cooling to the dew point. (I also just took the test and got this question right).

kompoz [17]4 years ago

4 0

The answer is b because as the air gets cooler and the moist in the air will drop as we call it dew

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What is the pOH of a solution that has a OH - concentration equal to 1.3x 10^-10

Paul [167]

Answer: 9.88 (approx10)

Explanation:

The pOH of a solution is the negative logarithm of the hydroxide-ion concentration

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3 years ago

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Need help ASAP... The diagram shows a box in the periodic table.. what is the atomic number of the element??

ad-work [718]

The atomic number is 15

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3 years ago

Consider the equation: S + 3O2 and SO3. Is this equation balanced? why or why not

Maurinko [17]

The balanced equation is
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3 years ago

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Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti

ella [17]

Answer :

The mass of excess mass of $Na_2CO_3$ , $AgNO_3,Ag_2CO_3\text{ and }NaNO_3$ are, 1.908 g, 0 g, 12.144 g and 3.74 g respectively.

Explanation : Given,

Mass of $Na_2CO_3$ = 4.25 g

Mass of $AgNO_3$ = 7.50 g

Molar mass of $Na_2CO_3$ = 106 g/mole

Molar mass of $AgNO_3$ = 170 g/mole

Molar mass of $Ag_2CO_3$ = 276 g/mole

Molar mass of $NaNO_3$ = 85 g/mole

First we have to calculate the moles of $Na_2CO_3$ and $AgNO_3$ .

$\text{Moles of }Na_2CO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Molar mass of }Na_2CO_3}=\frac{4.25g}{106g/mole}=0.040moles$

$\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{7.50g}{170g/mole}=0.044moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Na_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2NaNO_3$

From the balanced reaction we conclude that

As, 2 moles of $AgNO_3$ react with 1 mole of $Na_2CO_3$

So, 0.044 moles of $AgNO_3$ react with $\frac{0.044}{2}=0.022$ moles of $Na_2CO_3$

From this we conclude that, $Na_2CO_3$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

The excess mole of $Na_2CO_3$ = 0.040 - 0.022 = 0.018 mole

Now we have to calculate the mass of excess mole of $Na_2CO_3$ .

$\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3\times \text{Molar mass of }Na_2CO_3=(0.018mole)\times (106g/mole)=1.908g$

Now we have to calculate the moles of $Ag_2CO_3$ .

As, 1 moles of $AgNO_3$ react to give 1 moles of $Ag_2CO_3$

So, 0.044 moles of $AgNO_3$ react to give 0.044 moles of $Ag_2CO_3$

Now we have to calculate the mass of $AgCO_3$ .

$\text{Mass of }Ag_2CO_3=\text{Moles of }Ag_2CO_3\times \text{Molar mass of }Ag_2CO_3=(0.044mole)\times (276g/mole)=12.144g$

Now we have to calculate the moles of $NaNO_3$ .

As, 2 moles of $AgNO_3$ react to give 2 moles of $NaNO_3$

So, 0.044 moles of $AgNO_3$ react to give 0.044 moles of $NaNO_3$

Now we have to calculate the mass of $NaNO_3$ .

$\text{Mass of }NaNO_3=\text{Moles of }NaNO_3\times \text{Molar mass of }NaNO_3=(0.044mole)\times (85g/mole)=3.74g$

6 0

4 years ago

Which is a true statement about an exterior angle of a triangle?

wolverine [178]

The only true statement about an exterior angle of a triangle from the above is that it forms a linear pair with one of the interior angles of the triangle.

<h3>What is a triangle?</h3>

A triangle can be defined as a one of the plane shapes which has three sides and three angles

So therefore, the only true statement about an exterior angle of a triangle from the above is that it forms a linear pair with one of the interior angles of the triangle.

Learn more about triangles:

brainly.com/question/2217700

#SPJ1

7 0

2 years ago