Answer: 9.3 x 10^ 18 g CO
Explanation:
Start by knowing that carbon monoxide is the compound CO. To convert molecules to grams, you first need to convert molecules to moles. This can be done using the conversion factor for Avogadro's Number:
(2.0 x 10^5 molecules CO) x 1 mol CO / 6.02 x 10^23 molecules CO
This cancels molecules CO.
Then, you can convert moles to grams, which is your desired quantity. You can find the number of grams for CO by looking at the periodic table and adding together their masses. C = 12 g and O = 16 g. Total of 28 g CO:
(1 mol CO) x 28 g CO / 1 mol CO
This cancels mol CO, which leaves grams CO.
Answer:
i) The bond angle decreases due to the presence of lone pairs, which causes more repulsion on the bond pairs and as a result, the bond pairs tend to come closer. ii) The repulsion between electron pairs increases with an increase in electronegativity of the central atom and hence the bond angle increases.
Explanation:
Note the signs of equilibrium:-
- Reaction don't procede forward or backward
- Concentration of products and reactants remains same .
So
if
Concentration of A is 2M then concentration of B should be same .
So equilibrium constant K is 1
![\\ \rm\rightarrowtail K=\dfrac{[Products]^a}{[Reactants]^b}](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5Crightarrowtail%20K%3D%5Cdfrac%7B%5BProducts%5D%5Ea%7D%7B%5BReactants%5D%5Eb%7D)
So
Answer:
0.02 tons of NO produced when 4 tons of coal is burned
Explanation:
From the given,
Chemical formula of coal = 
Molecular mass of coal = 
Let’s calculate the mass of nitrogen in coal


Amount of coal burnt = 4tons
Amount of NO produced by burning 4tons = 
This nitrogen is converted into NO by reacting with atmospheric oxygen.
<u>Answer:</u> The equilibrium concentration of water is 0.597 M
<u>Explanation:</u>
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For a general chemical reaction:

The expression for
is written as:
![K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
The concentration of pure solids and pure liquids are taken as 1 in the expression.
For the given chemical reaction:

The expression of
for above equation is:
![K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2S%5D%5E2%5Ctimes%20%5BO_2%5D%7D)
We are given:
![[H_2S]_{eq}=0.671M](https://tex.z-dn.net/?f=%5BH_2S%5D_%7Beq%7D%3D0.671M)
![[O_2]_{eq}=0.587M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D0.587M)

Putting values in above expression, we get:
![1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}](https://tex.z-dn.net/?f=1.35%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%280.671%29%5E2%5Ctimes%200.587%7D)
![[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M](https://tex.z-dn.net/?f=%5BH_2O%5D%3D%5Csqrt%7B%281.35%5Ctimes%200.671%5Ctimes%200.671%5Ctimes%200.587%29%7D%3D0.597M)
Hence, the equilibrium concentration of water is 0.597 M