Answer:
Explanation:
A reactant is a substance that is present at the start of a chemical reaction. The substance(s) to the right of the arrow are called products . A product is a substance that is present at the end of a chemical reaction.
h2 + 02 = h2o
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Answer : The rms speed of the molecules in a sample of 
gas at 300 K will be four times larger than the rms speed of 
molecules at the same temperature, and the ratio 
constant with increasing temperature.
Explanation :
Formula used for root mean square speed :
where,
= rms speed of the molecule
R = gas constant
T = temperature
M = molar mass of the gas
At constant temperature, the formula becomes,
And the formula for two gases will be,
Molar mass of = 32 g/mole
Molar mass of = 2 g/mole
Now put all the given values in the above formula, we get
Therefore, the rms speed of the molecules in a sample of gas at 300 K will be four times larger than the rms speed of molecules at the same temperature.
And the ratio constant with increasing temperature because rms speed depends only on the molar mass of the gases at same temperature.
Answer:
An element that is oxidized is a reducing agent, because the element loses electrons, and an element that is reduced is an oxidizing agent, because the element gains electrons.
Answer: No charge (0)
Explanation:
The atom has a proton of 22 and electron number of 19. This means the element has lost 3 electrons. Therefore, the net charge is +3.
So, if this atom gains 3 more electrons, the net charge would be zero (neutral).
The total number of electron would now be 22 which is the same as the proton number.
This implies the atom is now in an unreacted state or ground state.
Answer:
The answer to your question is: 17.26% of carbon
Explanation:
Data
CxHy = 0.2121 g
BaCO₃ = 0.6006 g
Molecular mass BaCO₃ = 137 + 12 + 48 = 197 g
Reaction
CO₂ + Ba(OH)₂ ⇒ BaCO₃ + H₂O
Process
1.- Find the amount of carbon in BaCO₃
197 g of BaCO₃ --------------- 12 g of Carbon
0.6006 g ---------------- x
x = (0.6006 x 12) / 197
x = 0.0366 g of carbon
2.- Calculate the percentage of carbon in the organic compound
0.2121 g of organic compound --------------- 100%
0.0366g -------------- x
x = (0.0366 x 100) / 0.2121
x = 17.26%
