Answer:
1) Maximun ammount of nitrogen gas: 
2) Limiting reagent: 
3) Ammount of excess reagent: 
Explanation:
<u>The reaction </u>

Moles of nitrogen monoxide
Molecular weight: 


Moles of hydrogen
Molecular weight: 


Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess
1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted


2) <u>Limiting reagent</u>:
3) <u>Ammount of excess reagent</u>:


Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
False
Alkaline is something we use everyday
Acid is something sour eg lemons
➡ ANSWER
☑ <em><u>C</u></em><em><u>.</u></em><em><u> </u></em><em><u>3.5 105</u></em><em><u> </u></em><em><u>Hz</u></em>
Answer:
The final mass of sample is 1.3 g.
Explanation:
Given data:
Half life of H-3 = 12.32 years
Amount left for 15.0 years = 3.02 g
Final amount = ?
Solution:
First all we will calculate the decay constant.
t₁/₂ = ln² /k
t₁/₂ =12.32 years
12.32 y = ln² /k
k = ln²/12.32 y
k = 0.05626 y⁻¹
Now we will find the original amount:
ln (A°/A) = Kt
ln (3.02 g/ A) = 0.05626 y⁻¹ × 15.0 y
ln (3.02 g/ A) = 0.8439
3.02 g/ A = e⁰°⁸⁴³⁹
3.02 g/ A = 2.33
A = 3.02 g/ 2.33
A = 1.3 g
The final mass of sample is 1.3 g.