Answer:
1. First
2. Third
3. Fourth
4.remain the same as
Explanation:
Given the reaction equation;
Rate= k[A] [B]^3
We can see that the order of reaction is first order with respect to reactant A and third order with respect to reactant B. This gives an overall fourth order reaction.
If the concentration of A is doubled and that of B is halved. The rate of reaction remains the same.
Answer:
- The answer is the concentration of an NaOH = 1.6 M
Explanation:
The most common way to solve this kind of problem is to use the formula
In your problem,
For NaOH
C₁ =?? v₁= 78.0 mL = 0.078 L
For H₂SO₄
C₁ =1.25 M v₁= 50.0 mL = 0.05 L
but you must note that for the reaction of NaOH with H₂SO₄
2 mol of NaOH raect with 1 mol H₂SO₄
So, by applying in above formula
- (C₁ * 0.078 L) = (2* 1.25 M * 0.05 L)
- C₁ = (2* 1.25 M * 0.05 L) / (0.078 L) = 1.6 M
<u>So, the answer is the concentration of an NaOH = 1.6 M</u>
A covalent bond is stronger than an ionic bond because the covalent bond doesn’t separate in water
Answer:
1.667L of a 0.30M BaCl₂ solution
Explanation:
<em>Of a 0.30M barium chloride, contains 500.0mmol of barium chloride.</em>
<em />
Molarity is an unit of concentration used in chemistry defined as the moles of solute present in 1 liter of solution.
In a 0.30M BaCl₂ solution there are 0.30 moles of BaCl₂ in 1 liter of solution.
Now, in your solution you have 500mmol of BaCl₂ = 0.500 moles of BaCl₂ (1000 mmol = 1 mol). Thus, 0.500 moles of BaCl₂ are present in:
0.500 moles * (1L / 0.30 moles) =
<h3>1.667L of a 0.30M BaCl₂ solution</h3>
