Answer:
0.1313 g.
Explanation:
- It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.
- Suppose that hydrogen behaves ideally and at STP conditions.
<u><em>Using cross multiplication:</em></u>
1.0 mol of hydrogen occupies → 22.4 L.
??? mol of hydrogen occupies → 1.47 L.
∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.
- Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:
<em>The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen</em> = (6.563 x 10⁻² mol)(2.0 g/mol) = <em>0.1313 g.</em>
Answer:
The law is observed in the given equation.
Explanation:
CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂
In order to find out if the law of conservative mass is followed, we need to <u>count how many atoms of each element are there in both sides of the equation</u>:
- Ca ⇒ 1 on the left, 1 on the right.
- C ⇒ 1 on the left, 1 on the right.
- O ⇒ 3 on the left, 3 on the right.
- H ⇒ 2 on the left, 2 on the right.
- Cl ⇒ 2 on the left, 2 on the right.
As the numbers for all elements involved are the same, the law is observed in the given equation.
Answer:
It should b KNO3
Explanation:
one Potassium (K) and three Nitrite (NO3)
Answer:
C. The potential energy change for a chemical reaction.
Explanation:
The reaction coordinate q illustrates, graphically, the energy changes during exothermic and endothermic reactions. This graphical representation of the energy changes in the course of a chemical reaction is known as reaction coordinates. A reaction coordinate is a graphical sequence of steps by which the reaction progresses from reactants through activated complexes to products. Reaction coordinates explain how far a reaction has proceeded towards the products or from the reactants.
From the images attached below, we can see the reaction coordinates in the reaction profiles.
Answer:
IV
Explanation:
The complete question is shown in the image attached.
Let us call to mind the fact that the SN1 mechanism involves the formation of carbocation in the rate determining step. The order of stability of cabocations is; tertiary > secondary > primary > methyl.
Hence, a tertiary alkyl halide is more likely to undergo nucleophilic substitution reaction by SN1 mechanism since it forms a more stable cabocation in the rate determining step.
Structure IV is a tertiary alkyl halide, hence it is more likely to undergo nucleophilic substitution reaction by SN1 mechanism.