Climate is the weather that occur over a long period in a particular place.
Answer:
4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium
Explanation:
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:
- HCl: 2 moles
- Na: 1 mole
- NaCl: 2 moles
- H₂: 1 mole
You know the following masses of each element:
- H: 1 g/mole
- Cl: 35.45 g/mole
- Na: 23 g/mole
So, the molar mass of each compound participating in the reaction is:
- HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
- Na: 23 g/mole
- NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
- H₂: 2* 1 g/mole= 2 g/mole
Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:
- HCl: 2 moles* 36.45 g/mole= 72.9 g
- Na: 1 mole* 23 g/mole= 23 g
- NaCl: 2 moles* 58.45 g/mole= 116.9 g
- H₂: 1 mole* 2 g/mole= 2 g
So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

mass of H₂= 4.8 g
<u><em>4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium</em></u>
The activation energy Ea can be related to rate constant (k) at temperature (T) through the equation:
ln(k2/k1) = Ea/R[1/T1 - 1/T2]
where :
k1 is the rate constant at temperature T1
k2 is the rate constant at temperature T2
R = gas constant = 8.314 J/K-mol
Given data:
k1 = 0.543 s-1; T1 = 25 C = 25+273 = 298 K
k2 = 6.47 s-1; T = 47 C = 47+273 = 320 K
ln(6.47/0.543) = Ea/8.314 [1/298 - 1/320]
2.478 = 2.774 *10^-5 Ea
Ea = 0.8934*10^5 J = 89.3 kJ
Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.
<h3>What is volume?</h3>
Volume is the area occupied by the substance and is the ratio of the mass to the density.
At STP, 1 mole of gas occupies 22.4 L of volume
Given,
Volume of ammonia reacted = 0.720 L
The combustion reaction is shown as,

From the stoichiometry of the reaction, it can be said that,
L of ammonia reacts with
L of oxygen gas.
So, 0.720 L of ammonia will react with:

Therefore, the volume of oxygen required is 900 mL.
Learn more about volume here:
brainly.com/question/14090111
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The answer for this would be 69.6