Answer:
Your strategy here will be to use the molar mass of potassium bromide,
KBr
, as a conversion factor to help you find the mass of three moles of this compound.
So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.
Potassium bromide is an ionic compound that is made up of potassium cations,
K
+
, and bromide anions,
Br
−
. Essentially, one formula unit of potassium bromide contains a potassium atom and a bromine atom.
Use the periodic table to find the molar masses of these two elements. You will find
For K:
M
M
=
39.0963 g mol
−
1
For Br:
M
M
=
79.904 g mol
−
1
To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements
M
M KBr
=
39.0963 g mol
−
1
+
79.904 g mol
−
1
≈
119 g mol
−
So, if one mole of potassium bromide has a mas of
119 g
m it follows that three moles will have a mass of
3
moles KBr
⋅
molar mass of KBr
119 g
1
mole KBr
=
357 g
You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs
mass of 3 moles of KBr
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
360 g
a
a
∣
∣
−−−−−−−−−
Explanation:
<em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em> </em><em>3</em><em>6</em><em>0</em><em> </em><em>g</em><em> </em>
Fission is similar to mitosis because they both involve splitting.
Answer: <span><span>d. mitosis</span></span>
The balanced equation for the above reaction is as follows;
3NO₂ + H₂O --> 2HNO₃ + NO
stoichiometry of NO₂ to NO is 3:1
molar volume is where 1 mol of any gas occupies a volume of 22.4 L
volume of gas is directly proportional to number of moles of gas.
therefore stoichiometry can be applied for volume as well.
volume ratio of NO₂ to NO is 3:1
volume of NO₂ reacted - 854 L
therefore volume of NO formed - 854 L /3 = 285 L
volume of NO formed - 285 L
<u>Answer:</u> The value of
for the given reaction is 1.435
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

Given mass of
= 9.2 g
Molar mass of
= 92 g/mol
Volume of solution = 0.50 L
Putting values in above equation, we get:

For the given chemical equation:

<u>Initial:</u> 0.20
<u>At eqllm:</u> 0.20-x 2x
We are given:
Equilibrium concentration of
= 0.057
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![[NO_2]_{eq}=2x=(2\times 0.143)=0.286M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.143%29%3D0.286M)
![[N_2O_4]_{eq}=0.057M](https://tex.z-dn.net/?f=%5BN_2O_4%5D_%7Beq%7D%3D0.057M)
Putting values in above expression, we get:

Hence, the value of
for the given reaction is 1.435
Answer:
C. Atomic mass
Explanation:
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