The answer is C.
The vast difference in electronegativity of the oxygen and hydrogen in water, the O-H bond is polar.
Answer:
6
Explanation:
the value is 6 because its an even number
Answer:
1 mol
Explanation:
Using the general gas law equation as follows:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the provided information in the question;
V = 22.4L
T = 273K
P = 1 atm
R = 0.0821 Latm/molK
n = ?
Using PV = nRT
n = PV/RT
n = (1 × 22.4) ÷ (0.0821 × 273)
n = 22.4 ÷ 22.4
n = 1mol
Answer:
0.144M
Explanation:
First, let us write a balanced equation for the reaction. This is illustrated below:
HNO3 + KOH —> KNO3 + H20
From the equation,
nA = 1
nB = 1
From the question given, we obtained the following:
Ma =?
Va = 30.00mL
Mb = 0.1000M
Vb = 43.13 mL
MaVa / MbVb = nA/nB
Ma x 30 / 0.1 x 43.13 = 1
Cross multiply to express in linear form
Ma x 30 = 0.1 x 43.13
Divide both side by 30
Ma = (0.1 x 43.13) /30 = 0.144M
The molarity of the nitric acid is 0.144M
<u>Answer:</u> The equilibrium concentration of bromine gas is 0.00135 M
<u>Explanation:</u>
We are given:
Initial concentration of chlorine gas = 0.0300 M
Initial concentration of bromine monochlorine = 0.0200 M
For the given chemical equation:
<u>Initial:</u> 0.02 0.03
<u>At eqllm:</u> 0.02-2x x 0.03+x
The expression of 
for above equation follows:
![K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BBr_2%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BBrCl%5D%5E2%7D)
We are given:
Putting values in above equation, we get:
Neglecting the value of x = -0.96 because, concentration cannot be negative
So, equilibrium concentration of bromine gas = x = 0.00135 M
Hence, the equilibrium concentration of bromine gas is 0.00135 M
