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Fofino [41]
3 years ago
13

What effects does the concentration of reactants have on the rate of a reaction ?

Chemistry
1 answer:
Harlamova29_29 [7]3 years ago
5 0

Question:

<em>What effects does the concentration of reactants have on the rate of a reaction?</em>

Answer:

<em>Reactant concentration. Increasing the concentration of one or more reactants will often increase the rate of reaction. This occurs because a higher concentration of a reactant will lead to more collisions of that reactant in a specific time period.</em>

<em>Increasing the concentration of reactants generally increases the rate of reaction because more of the reacting molecules or ions are present to form the reaction products. ... When concentrations are already high, a limit is often reached where increasing the concentration has little effect on the rate of reaction.</em>

Hope this helps, have a good day. c;


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8 0
3 years ago
The most common isotopic forms of hydrogen are ordinary hydrogen (1H) and deuterium (2H), which have percent compositions of 99.
Alborosie

Answer:

0. 000115

Explanation:

A percentage is defined as a ratio with a basis of 100 as total substance. Convert a percentage to decimal implies to divide the percentage in 100 because decimal form has as basis 1.

For the isotopic forms:

1H: 99.98% → As percent.

99.98% / 100 = 0.9998 → As decimal form.

2H: 0.0115% → As percent.

0.0115% / 100 = <em>0. 000115</em>→ As decimal form.

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5 0
3 years ago
What is the concentration of bromide, in ppm, if 12.41 g MgBr2 is dissolved in 2.55 L water.
pav-90 [236]

Answer:

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

Explanation:

ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.

In our case we have:

mass of MgBr₂ = 12.41 g

volume of water (which is equal to the final solution volume) = 2.55 L

Now we devise the following reasoning:

if         12.41 g of MgBr₂ are dissolved in 2.55 L of water

then         X g of MgBr₂ are dissolved in 1 L of water

X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂

if in         184 g (1 mole) of MgBr₂ we have 160 g of Br⁻

then in   4.867 g of MgBr₂ we have Y g of Br⁻

Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)

4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

7 0
4 years ago
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