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3 years ago

Which energy change occurs during boiling?

Chemistry

2 answers:

faltersainse [42]3 years ago

8 0

Answer:

heat energy is absorbed by the substance (4th option)

Explanation:

i took the test & got this question correct.

dybincka [34]3 years ago

5 0

Answer:

heat energy because u are boiling something

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Which of these statement(s) is/are true about a balloon filled with 1.00 mol N2 (g) at STP?

lara [203]

Answer:

I. The balloon has a volume of 22.4L

III. The balloon contains 6.022x10^23 molecules.

Explanation:

At stp, it has been proven that 1mole of a gas occupy 22.4L.

Therefore, option (i) is correct.

The molar mass N2 = 14.01 x 2 = 28.02g/mol

Number of mole of N2 = 1 mole

Mass of N2 =..?

Mass = mole x molar Mass

Mass of N2 = 1 x 28.02 = 28.02g.

The mass content of the balloon is 28.02g, therefore, option (ii) is wrong.

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 molecules. This implies that 1 mole of N2 also contains 6.02x10^23 molecules

Therefore, option (iii) is correct.

The correct options to the question are:

Option i and option iii

7 0

3 years ago

Read 2 more answers

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

How many grams are in a 1.48x107 ig

Firdavs [7]

Answer:978989989.1429323

Explanation:

6 0

2 years ago

10. When a piece of steel wool is burned it gains mass. Which explanation is consistent with the law of conservation of mass?

frez [133]

Answer:

Oh come on. Look at all - then look at A.

Explanation:

4 0

3 years ago

The first-order rate constant for the decomposition of n2o5, 2n2o5(g)→4no2(g)+o2(g) at 70∘c is 6.82×10−3 s−1. suppose we start w

torisob [31]

The rate constant for 1st order reaction is

K = (2.303 /t) log (A0 /A)

Where, k is rate constant

t is time in sec

A0 is initial concentration

(6.82 * 10-3) * 240 = log (0.02 /A)

1.63 = log (0.02 /A)

-1.69 – log A = 1.63

Log A = - 0.069

A = 0.82

Hence, 0.82 mol of A remain after 4 minutes.

7 0

3 years ago