Answer:answers are in the explanation
Explanation:
(a). pH less than 7 between 1 - 3.5 are strong acid, and between 4.5-6.9 weak acid.
pH greater than 7; between 10-14 is a strong base, and between 7.1 - 9, it is weakly basic.
(b). Equation of reaction;
HBr + KOH ---------> KBr + H2O
One mole of HBr reacts with one mole of KOH to give one Mole of KBr and one mole of H2O
Calculating the mmol, we have;
mmol KOH = 28.0 ml × 0.50 M
mmol KOH= 14 mmol
mmol of HBr= 56 ml × 0.25M
mmol of HBr= 14 mmol
Both HBr and KOH are used up in the reaction, which leaves only the product,KBr and H2O.
The pH here is greater than 7
(C). [NH4^+] = 0.20 mol L^-1 × 50 ml. L^-1 ÷ 50 mL + 50mL
= 0.10 M
Ka=Kw/kb
10^-14/ 1.8× 10^-5
Ka= 5.56 ×10^-10
Therefore, ka= x^2 / 0.20
5.56e-10 = x^2/0.20
x= (0.20 × 5.56e-10)^2
x= 1.05 × 10^-5
pH = -log [H+]
pH= - log[1.05 × 10^-5]
pH = 4.98
Acidic(less than 7)
(c). 0.5 × 20/40
= 0.25 M
Ka= Kw/kb
kb= 10^-14/1.8× 10^-5
Kb = 5.56×10^-10
x= (5.56×10^-10 × 0.5)^2
x= 1.667×10^-5 M
pH will be basic
Answer:
See explanation
Explanation:
Step 1: Data given
For the reaction aA + bB ⇆ cC + dD
Kc = [C]^c * [D]^d / [A]^a [B]^b
a. SbCl5(g) ⇄ SbCl3(g) + Cl2(g)
Kc = [Cl2]*[SbCl3] / [SbCl5]
b. 2 BrNO(g) ⇄ 2NO(g) + Br2(g)
Kc = [Br2]*[NO]² / [BrNO]²
c. CH4(g) + 2 H2S(g) ⇄ CS2(g) + 4 H2(g)
Kc = [H2]^41 * [CS2] / [H2S]²*[CH4]
d. 2CO(g) + O2(g) ⇄ 2CO2(g)
Kc = [CO2]² / [O2][CO]²
Answer is 0.289nm.
Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.
wt % of Fe in Fe-V alloy = 85%
wt % of V in Fe-V alloy = 15%
We need to calculate edge length of the unit cell having bcc structure.
Using density formula,
For calculating edge length,
For calculating 
, we use the formula
Similarly for calculating 
, we use the formula
From the periodic table, masses of the two elements can be written
Specific density of both the elements are
Putting and 
formula's in edge length formula, we get
![a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}](https://tex.z-dn.net/?f=a%3D%5Cleft%20%5B%5Cfrac%7BZ%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B%28wt%5C%25%29_%7BFe%7D%7D%7BM_%7BFe%7D%7D%2B%5Cfrac%7B%28wt%5C%25%29_%7BFe%7D%7D%7BM_%7BFe%7D%7D%7D%20%20%5Cright%20%29%7D%7BN_A%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B%28wt%5C%25%29_V%7D%7B%5Crho_V%7D%2B%5Cfrac%7B%28wt%5C%25%29_V%7D%7B%5Crho_V%7D%7D%20%20%5Cright%20%29%7D%20%20%5Cright%20%5D%5E%7B1%2F3%7D)
![a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}](https://tex.z-dn.net/?f=a%3D%5Cleft%20%5B%5Cfrac%7B2atoms%2F%5Ctext%7Bunit%20cell%7D%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B85%5C%25%7D%7B55.85g%2Fmol%7D%2B%5Cfrac%7B15%5C%25%7D%7B50.941g%2Fmol%7D%7D%20%20%5Cright%20%29%7D%7B%286.023%5Ctimes10%5E%7B23%7Datoms%2Fmol%29%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B85%5C%25%7D%7B7.874g%2Fcm%5E3%7D%2B%5Cfrac%7B15%5C%25%7D%7B6.10g%2Fcm%5E3%7D%7D%20%20%5Cright%20%29%7D%20%20%5Cright%20%5D%5E%7B1%2F3%7D)
By calculating, we get
Answer: An object has a weight of 10 kg on the surface of Earth. If the same object were transported to the surface of Mars, the object would have a weight of 3.8 kg. The density of the object is greater on Earth than it is on Mars.
Explanation:
The given question is an incomplete question, the complete question is given below.
Use this equation to answer the questions that follow.
5 moles of oxygen gas are present. How many moles of 
will be produced with this amount of oxygen gas?
Answer : The moles of produced will be 3.33 moles.
Explanation : Given,
Moles of oxygen gas = 5 moles
Now we have to calculate the moles of .
The given balanced chemical equation is:
From the balanced chemical equation we conclude that,
As, 3 moles of 
gas react to produces 2 moles of gas
So, 5 moles of gas react to produces 
moles of gas
Therefore, the moles of produced will be 3.33 moles.
