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Alexeev081 [22]

3 years ago

12

How many particles of BaCl2 reacted if 1.53 mol of it were consumed?

1 answer:

sergejj [24]3 years ago

3 0

Answer:- D) $9.21*10^2^3$ particles $BaCl_2$ .

Solution:- There are 1.53 moles of the compound are given and it asks to calculate the number of particles.

To calculate the number of particles we multiply the moles with Avogadro number.

$1.53mol(\frac{6.022*10^2^3particles}{1mol})$

= $9.21*10^2^3$

So, the D is the right choice.

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3.0 cm x 4.0 cm x 1.0 cm<br><br>[?]cm^3

CaHeK987 [17]

Explanation:

<em>Hi</em><em> </em><em>there</em><em>!</em><em>!</em>

<em>you</em><em> </em><em>asked</em><em> </em><em>to</em><em> </em><em>multiply</em><em> </em><em>these</em><em> </em><em>all</em><em> </em><em>right</em><em>,</em>

<em>you</em><em> </em><em>can</em><em> </em><em>simply</em><em> </em><em>multiply</em><em> </em><em>it</em><em> </em><em>,</em>

<em>=</em><em>3</em><em>cm</em><em> </em><em>×</em><em> </em><em>4</em><em> </em><em>cm</em><em> </em><em>×</em><em> </em><em>1</em><em>cm</em>

<em>=</em><em> </em><em>1</em><em>2</em><em>cm</em><em>^</em><em>2</em><em>×</em><em>1</em><em>cm</em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>4</em><em>×</em><em>3</em><em>=</em><em>1</em><em>2</em><em>)</em>

<em>=</em><em> </em><em>1</em><em>2</em><em>cm</em><em>^</em><em>3</em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>1</em><em>2</em><em>×</em><em>1</em><em>=</em><em>1</em><em>2</em><em>)</em>

<em>Therefore</em><em>, </em><em> </em><em>the</em><em>answer is</em><em> </em><em>1</em><em>2</em><em> </em><em>cm</em><em>^</em><em>3</em><em>.</em>

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em>

4 0

4 years ago

A carbon atom can form four covalent bonds. How many valence electrons does it have?

pshichka [43]

If a carbon atom can form four covalent bonds, it also has four valence electrons. For neutral atoms, the number of valence electrons the same as an atom's main group number.

8 0

3 years ago

Calculate the unknown quantity in each of the following measurements of gases ?

hjlf

You have to google the answers

5 0

4 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

Why is it important for scientists to describe their procedures in full detail?

Licemer1 [7]

So that other scientist can repeat their experiments
i hope i got i right

6 0

3 years ago