Answer:
1 : 1.5
Explanation:
First Sample;
Ratio of sulfur and Oxygen
Mass of sulfur : Mass of oxygen
Mass of oxygen = Mass of sample - Mass of sulfur = 70 - 35 = 35g
35g : 35g
1 : 1
Second Sample;
Ratio of sulfur and Oxygen
Mass of sulfur : Mass of oxygen
Mass of oxygen = Mass of sample - Mass of sulfur = 70 - 28 = 42g
28g : 42g
1 : 1.5
Further reducing it to make oxygen 1;
0.6667 : 1
ratio in whole numbers of the masses of sulfur that combine with 1.00 g of oxygen between the two compounds;
0.6667 : 1
1 : 1.5
Answer:
Answer: Chemical change; Chemical property of heat of combustion
Explanation:
A physical change is defined as a change in which there is alteration in shape, size etc. No new substance gets formed in these reactions.
Explanation:
To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows
time 0 40 80 120 160
moles 0.100 0.067 0.045 0.030 0.020
Q1)
for the first 40 s change of moles ;
= -d[A] / t
= - (0.067-0.100)/40s
= 8.25 x 10⁻⁴ mol/s
for the next 40 s
= -(0.045-0.067)/40
= 5.5 x 10⁻⁴ mol/s
the 40 s after that
= -(0.030-0.045)/40 s
= 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction
rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction
Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation.
Since this is a first order reaction,
b = 1
therefore the reaction is
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
= 8.25 x 10⁻³ s⁻¹
Answer:
The precipitated are option a and d.
Explanation:
2 LiI(aq) +Hg2(NO3)2(aq) → Hg2I2(s) ↓ + 2 LiNO3(aq)
Cation Hg2+ 2 in the presence of iodide, a precipitated is formed.
Zn(s) + 2AgNO3(aq) → 2 Ag(s) ↓ +Zn(NO3)2(aq)
Zinc starts to get rid, and some white particles also stick to it. Afterwards the solution becomes cloudy and a precipitate appears, which is the solid silver
To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:
Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]
HOBr = 0.50 M
KOBr = 0.30 M = OBr-
<span> HOBr + H2O <-> H+ + OBr- </span>
<span>I 0.50 - 0 0.30 </span>
<span>C -x x x
</span>---------------------------------------------
<span>E(0.50-x) x (0.30+x) </span>
<span>Assuming that the value of x is small as compared to 0.30 and 0.50 </span>
<span>Ka = 2.0 x 10^-9 = x (0.30) / 0.50) </span>
<span>x = 3.33 x 10^-9 = H+</span>
pH = 8.48
