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Oduvanchick [21]
3 years ago
10

How many liters of 1.2M solution can be prepared with 0.50 moles of C6H12O6

Chemistry
1 answer:
BigorU [14]3 years ago
5 0
First you need to know the molecular weight of sugar (C6H12O6) which is 180.156g/mol
You have half a mole so you have 90.078g
If you wanted to make 1L of a 1.2M solution of glucose you would need 180.156*1.2=216.1872g
But you only have 90.078g
So you need to figure out how much this 90.078g will make if the solution must be 1.2M:
90.078g/216.1872g=xL/1L
solve for the X and you get 0.416666666...
so 416.7ml or 0.417L
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sertanlavr [38]

hydrogen combines with sulfur is H2S

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6 0
3 years ago
Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it
lions [1.4K]

<u>Answer:</u> The equilibrium concentration of COF_2 is 0.332 M

<u>Explanation:</u>

We are given:

Initial concentration of COF_2 = 2.00 M

The given chemical equation follows:

                2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)

<u>Initial:</u>          2.00

<u>At eqllm:</u>     2.00-2x          x      x

The expression of K_c for above equation follows:

K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}

We are given:

K_c=6.30

Putting values in above expression, we get:

6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M

Hence, the equilibrium concentration of COF_2 is 0.332 M

4 0
3 years ago
Which environment can occur for a short while and usually found in low level areas along rivers, lakes, and streams?
Svetradugi [14.3K]
The answer is wetland
3 0
3 years ago
Read 2 more answers
What is the wavelength of a wave?
liberstina [14]

Answer:

Wavelength (λ) – The distance of one complete cycle in the wave. The distance between two consecutive crests and /or troughs. S.I. Unit: metre (m).

In other words:

Wavelength is the distance from crest to crest (or trough to trough).

5 0
3 years ago
A 10.0 mL sample of HNO3 was exactly neutralized by 13.5 mL of 1.0 M KOH. What is the molarity of the HNO3? Use the titrations f
Kruka [31]

Answer: Thus molarity of HNO_3 is 1.35 M

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HNO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=1\\M_1=?M\\V_1=10.0mL\\n_2=1\\M_2=1.0M\\V_2=13.5mL

Putting values in above equation, we get:

1\times M_1\times 10.0=1\times 1.0\times 13.5\\\\M_1=1.35M

Thus molarity of HNO_3 is 1.35 M

7 0
3 years ago
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