Answer is: 1,92 mol/L·s.
Chemical reaction: 2D(g) + 3E(g) + F(g) → <span>2G(g) + H(g).
</span>H is increasing at 0,64 mol/L·<span>s.
From chemical reaction n(H) : n(E) = 1 : 3.
0,64 mol : n(E) = 1 : 3.
n(E) = 1,92 mol.
</span>E is decreasing at 1,92 mol/L·s.
The question is: Give the systematic name of 
.
Answer: The systematic name of is silicon difluoride.
Explanation:
The given compound has chemical formula . It shows that there is one atom of silicon and two atoms of fluorine are present.
So, the number "two" will be represented by the prefix "di" while naming this compound.
Hence, systematic name of this compound is silicon difluoride.
Thus, we can conclude that systematic name of is silicon difluoride.
Answer:
4 × 10 g
Explanation:
Step 1: Write the balanced equation
2 H₂(g) + O₂(g) ⇒ 2 H₂O(I)
Step 2: Calculate the moles corresponding to 4 g of H₂
The molar mass of H₂ is 2.02 g/mol.
4 g × 1 mol/2.02 g = 2 mol
Step 3: Calculate the moles of H₂O produced from 2 moles of H₂
The molar ratio of H₂ to H₂O is 2:2. The moles of H₂O produced are 2/2 × 2 mol = 2 mol.
Step 4: Calculate the mass corresponding to 2 moles of H₂O
The molar mass of H₂O is 18.02 g/mol.
2 mol × 18.02 g/mol = 4 × 10 g
Answer:
(C) It is favorable and is driven by ΔH° only.
Explanation:
Let's consider the following balanced equation.
2 Na₂O₂ + S + 2 H₂O → 4 NaOH + SO₂
To determine whether it will be favorable or not at 298 K, we need to calculate the standard Gibbs free energy (ΔG°).
- If ΔG° < 0, the reaction will be favorable.
- If ΔG° > 0, the reaction will be unfavorable.
We can calculate ΔG° using the following expression.
ΔG° = ΔH° - T.ΔS°
As we can see from the expression above, the favorability will be driven if ΔH° < 0 and if ΔS° < 0. Since ΔH° = -610 kJ/mol and ΔS° = -73 J/K.mol, the favorability will be driven by ΔH°. Now, let's calculate the overall favorability.
ΔG° = ΔH° - T.ΔS°
ΔG° = -610 kJ/mol - 298 K.(-0.073 kJ/K.mol) = -588 kJ/mol
The reaction is favorable and is driven by ΔH° only.
I would say carbon monoxide
Firstly because the image shows two different atoms bonded so it cannot be nitrogen or krypton
Secondly sulphur dioxide has 3 atoms bonded (two oxygens and a sulphur atom) so it can’t be that
Finally it can’t be hydrogen chloride because chloride is significantly larger than hydrogen
Thus it must be carbon monoxide as carbon and oxygen are bonded (CO) and are both relatively similar in size
