Answer:
See explanation
Explanation:
Copper II oxide is a base but not an alkali. An alkali is a soluble base. Since Copper II oxide is not soluble in water then it is not an alkali.
Let us recall that the change of colour of litmus with an alkali requires the presence of water. In the absence of water, solid Copper II oxide does not turn red litmus paper blue.
The ability to turn red litmus paper blue is commonly observed with alkalis and Copper II oxide is not an alkali.
Also recall that since Copper II oxide is not soluble, hydroxide ions are absent hence Copper II oxide does not turn red litmus paper blue.
Answer:
NaCl + Fe2+ ---> FeCl2 + Na+
Single replacement
Explanation:
Na is replaced by Fe2+ to form FeCl2, and that is the only replacement. Thus, it is single replacement.
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>:
The answer is: 27 grams of aluminium.
Balanced chemical reaction: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.
n(H₂) = 1.5 mol; amount of hydrogen.
Form chemical reaction: n(Al) : n(H₂) = 2 : 3.
n(Al) = 2 · 1.5 mol ÷ 3.
n(Al) = 1.0 mol; amount of aluminium.
m(Al) = n(Al) · M(Al).
m(Al) = 1 mol · 27 g/mol.
m(Al) = 27 g; mass of aluminium.
Answer: 4.41 atm
Explanation:
Given that,
Original pressure of oxygen gas (P1) = 5.00 atm
Original temperature of oxygen gas (T1) = 25°C
[Convert 25°C to Kelvin by adding 273
25°C + 273 = 298K
New pressure of oxygen gas (P2) = ?
New temperature of oxygen gas (T2) = -10°C
[Convert -10°C to Kelvin by adding 273
-10°C + 273 = 263K
Since pressure and temperature are given while volume is held constant, apply the formula for Charle's law
P1/T1 = P2/T2
5.00 atm /298K = P2/263K
To get the value of P2, cross multiply
5.00 atm x 263K = 298K x V2
1315 atm•K = 298K•V2
V2 = 1315 atm•K / 298K
V2 = 4.41 atm
Thus, the new pressure inside the canister is 4.41 atmosphere
