Answer:

Explanation:
The limiting reactant is the reactant that gives the smaller amount of product.
Assemble all the data in one place, with molar masses above the formulas and masses below them.
M_r: 39.10 80.41 2.016
2K + 2HBr ⟶ 2KBr + H₂
m/g: 5.5 4.04
a) Limiting reactant
(i) Calculate the moles of each reactant

(ii) Calculate the moles of H₂ we can obtain from each reactant.
From K:
The molar ratio of H₂:K is 1:2.

From HBr:
The molar ratio of H₂:HBr is 3:2.

(iii) Identify the limiting reactant
HBr is the limiting reactant because it gives the smaller amount of NH₃.
b) Excess reactant
The excess reactant is K.
c) Mass of H₂

Answer:
in days
Explanation:
A short term environmental change refers to changes in the environment that happens within a short period of time, for example, flooding, drought, volcanic eruptions etc. In contrast to short term environmental change, long term environmental change takes place over a long period of time hence species have a longer time to adapt to it. Short term environmental changes may or may not be caused by humans. An example of a short term environmental change caused by humans is oil spillage. Short term environmental changes may lead to the the migration of species from the ecosystem.
Short term environmental changes usually occur within a very short time. The greatest time frame in which a short term environmental change can occur is in a matter of days. Only longterm environmental changes span hundreds and thousands of years.
Answer:
Problem, Variables, Hypothesis, Research, Procedure, Data, Conclusion.
Explanation:
First, you have to know what the problem that you're talking about is. Then, you have to know the independent, dependent, and controlled variables are in order to make a hypothesis. Then do research about the topic. After that, you make the procedure your experiment, if that's what your doing. Then after the experiment, you find the data. Using that data, you make a conclusion.
Answer:
1.3 × 10⁴ mL
Explanation:
A Pb-contaminated water sample contains 0.0012 %Pb by mass, that is, there are 0.0012 g of Pb in 100 g of solution. The mass of the sample that contains 150 mg (0.150 g) of Pb is:
0.150 g Pb × (100 g sample/0.0012 g Pb) = 1.25 × 10⁴ g sample
The density of the sample is 1.0 g/mL. The volume of the sample is:
1.25 × 10⁴ g × (1 mL/1.0 g) = 1.3 × 10⁴ mL