Answer:
6.9 ml of concentrate
Explanation:
100 ml of .1 M will require .01 moles
from a 1.45 M solution, .01 mole would be
.01 mole / ( 1.45 mole / liter) = 6.9 ml of the concentrate then dilute to 100 ml
Answer:
a) Limiting: sulfur. Excess: aluminium.
b) 1.56g Al₂S₃.
c) 0.72g Al
Explanation:
Hello,
In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.
b) By stoichiometry, the produced grams of aluminium sulfide are:

c) The leftover is computed as follows:

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.
Best regards.
The most likely answer is the boiling point and freezing point of water. The Celsius scale starts at the freezing point of water (0°C) and than scaled so that 100°C fell on the boiling point of water.
I hope this helps. Let me know if anything is unclear.
Carbon Dioxide is formed during metabolic reactions of cellular respiration.
Answer:
0.57 water
Explanation:
To solve this problem, we need to write the reaction expression first.
The reactants are oxygen gas and hydrogen gas.
They react to give a product of water
2H₂ + O₂ → 2 H₂O
Given that;
Number of moles of hydrogen gas = 0.57moles
From the balanced reaction expression;
2 moles of hydrogen gas produces 2 moles of water
So;
0.57mole of hydrogen gas will also produce 0.57 water