Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
Yes. Exhale into a jar.
That's a hella ratchet way to capture CO2, but it works nonetheless
Answer:
0.125. work- divide the volume value by 1000
873. work- multiply the length value by 100
98100. work- Conversion factor: 1 kg = 100000 cg
1) Centigram = Kilogram * 100000
2) Centigram = 0.981 * 100000
3) Centigram = 98100
285.65. work- 12.5°C + 273.15 = 285.65K
446.85. work- 720K − 273.15 = 446.85°C
346.25. work- 73.1°C + 273.15 = 346.25K
Explanation:
i hope this helps:) brainliest plss??
Answer:
a. 59 m/atm
Explanation:
- To solve this problem, we must mention Henry's law.
- <em>Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.</em>
- It can be expressed as: C = KP,
C is the concentration of the solution (C = 1.3 M).
P is the partial pressure of the gas above the solution (P = 0.022 atm).
K is the Henry's law constant (K = ??? M/atm),
∵ C = KP.
∴ K = C/P = (1.3 M)/(0.022 atm) = 59.0 M/atm.
