Answer:
Zn3P2O8
Explanation:
In this particular question, it is necessary to convert the respective masses to percentages. We convert to percentages by placing each mass over the total mass and multiplying by 100%. Since the total is 50mg, conversion to percentage can be done by multiplying the masses by 2 as 100/50 is 2
For Oxygen = 16.58 * 2 = 33.16%
For phosphorus = 8.02 * 2 = 16.04%
For zinc = 25.40 * 2 = 50.80%
We then proceed to divide these percentages by their respective atomic masses. The atomic mass of oxygen, phosphorus and zinc are 16, 31 and 65 respectively.
O = 33.16/16 = 2.0725
P = 16.04/31 = 0.5174
Zn = 50.80/65 = 0.7815
Now, we divide by the smallest value which is that of the phosphorus
O = 2.0725/0.5174 = 4
P = 0.5174/0.5174 = 1
Zn= 0.7815/0.5174 = 1.5
Now, we need to multiply through by 2. This yields: O = 8, P = 2 and Zn = 3
The empirical formula is thus: Zn3P2O8
Answer:
c = 0.07 j/g.k
Explanation:
Given data:
Mass of sample = 35 g
Heat absorbed = 48 j
Initial temperature = 293 K
Final temperature = 313 K
Specific heat of substance = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
ΔT = 313 k - 293 K
ΔT = 20 k
Now we will put the values in formula.
48 j = 35 g × c× 20 k
48 j = 700 g.k ×c
c = 48 j/700 g.k
c = 0.07 j/g.k
Answer:
8.08 × 10⁻⁴
Explanation:
Let's consider the following reaction.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
The initial concentration of phosgene is:
M = 2.00 mol / 1.00 L = 2.00 M
We can find the final concentrations using an ICE chart.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
I 2.00 0 0
C -x +x +x
E 2.00 -x x x
The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.
The concentrations at equilibrium are:
[COCl₂] = 2.00 -x = 1.96 M
[CO] = [Cl₂] = 0.0398 M
The equilibrium constant (Keq) is:
Keq = [CO].[Cl₂]/[COCl₂]
Keq = (0.0398)²/1.96
Keq = 8.08 × 10⁻⁴
Answer:
Most liking the puck will go flying because of the force of the hockey stick.
Answer: The answer is A. A conductor that allows electricity to flow easily
