Word equation for k(s)+H2O(l) yield KOH(aq) + H2(g)
2 answers:
<u>Answer:</u> The word equation for the given chemical equation is given below.
<u>Explanation:</u>
Word equation is defined as the equation in which substances are written in word form and not in chemical formulas. The words "and" or "plus" means that one substance and another substance are both reactants or products.
For the given chemical equation:
The word equation for this follows:
Here, 1 mole of potassium metal reacts with liquid water to produce 1 mole of potassium hydroxide and 1 mole of hydrogen gas.
Hence, the word equation for the given chemical equation is given above.
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Answer: A) 0.0480 grams and B) 56.16 years.
Explanation: Half live is the time in which the amount of radioactive substance remains halve of its initial amount.
The formula we use for solving this type of problem is:
where, is the initial amount and N is the remaining amount of radioactive substance and n is the number of half lives.
where, T is the time and is half life.
A) from given data, = 2 g
T = 70 years
= 13 years
N= ?
n = 5.38
N = 0.0480 g
So, 0.0480 grams of the substance will be remaining after 70 years.
B) = 2 g
N = 0.1 g
T = ?
Let's first calculate the value of n for this.
Taking log to both sides:
n = 4.32
Half life is 13 years, so we can calculate the time as:
T = 56.16 years
So, it will take 56.16 years for the radioactive substance to decay from 2 g to 0.1 g.
Answer:
Oxygen is the limiting reactant.
Explanation:
Hello,
In this case, given the reaction:
Hence, given the masses of both ethanol and oxygen, we are able to compute the available moles ethanol by:
Next, we compute the moles of ethanol that react with the 0.640 grams of oxygen considering their 1:3 molar ratio in the chemical reaction:
In such a way, since there are 0.01 available moles of ethanol but just 0.0067 moles are reacting, we evidence ethanol is in excess, therefore the oxygen is the limiting reactant.
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