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tangare [24]
3 years ago
7

How many moles of KOH are measured in thisexperiment? See the volume and molarity information given. 2.How many moles of KOH rea

cted with the amount of Al you used? Refer to the reaction equation in the first few lines of the experiment.3.How many moles of KOH were left over?

Chemistry
1 answer:
Daniel [21]3 years ago
5 0

Answer:

0.022 moles of potassium hydroxide reacted with Aluminum and 0.013 moles of potassium hydroxide were left over.

Explanation:

2Al(s) + 2KOH(aq) + 6H2O(l) → 2KAl(OH)4(aq) + 3H2(g)

First we obtain the amount in moles of aluminum added in line with the mass of aluminum to be added in the experimental procedure. This shows us that aluminum metal is the limiting reactant.

Secondly we obtain the amount of KOH in 25mL of 1.4M KOH solution(see attached image). This gives the amount of KOH availiable for reaction.

2 moles of Aluminum metal reacts with 2 moles of KOH according to the balanced reaction equation. Hence 0.022 moles of aluminum will react with 0.022 moles of KOH.

The amount of KOH left unreacted is 0.035-0.022= 0.013 moles of KOH.

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What volume of 0.305 m agno3 is required to react exactly with 155.0 ml of 0.274 m na2so4 solution? hint: you will want to write
LuckyWell [14K]

The balanced chemical equation for reaction of AgNO_{3} and Na_{2}SO_{4} is as follows:

2 AgNO_{3}+Na_{2}SO_{4}\rightarrow 2NaNO_{3}+Ag_{2}SO_{4}

From the balanced chemical equation, 2 mol of AgNO_{3} reacts with 1 mol of  NaNO_{3}.

First calculating number of moles of NaNO_{3} as follows:

M=\frac{n}{V}

On rearranging,

n=M\times V

Here, M is molarity and V is volume. The molarity of NaNO_{3}  is given 0.274 M or mol/L and volume 155 mL, putting the values,

n=0.274 mol/L\times 155\times 10^{-3}mL=0.04247 mol

Since, 1 mol of NaNO_{3}  reacts with 2 mol of  AgNO_{3} thus, number of moles of  AgNO_{3}  will be 2\times 0.04247 mol=0.08494 mol.

Now, molarity of  AgNO_{3} is given 0.305 M or mol/L thus, volume can be calculated as follows:

V=\frac{n}{M}=\frac{0.08494 mol}{0.305 mol/L}=0.2785 L=278.5 mL

Therefore, volume of  AgNO_{3} is 278.5 mL.

4 0
3 years ago
In a titration, 45.0 mL of KOH is neutralized by 75.0 mL of 0.30M HBr. How much KOH is in 1.0 liter of the KOH solution ?
photoshop1234 [79]
KOH + HBr ---> KBr + H2O

0,3 moles of HBr ---in-------1000ml
x moles of HBr-------in------75ml
x = 0,0225 moles of HBr

according to the reaction:  1 mole of KOH = 1 mole of HBr
so
0,0225 moles of HBr = 0,0225 moles of KOH

0,0225 mole of KOH------in-----45ml
x moles of KOH -----------in------1000ml
x = 0,5 moles of KOH

answer: 0,5 mol/dm³  KOH (molarity)


8 0
3 years ago
-The picture goes to Activity 2
Alborosie

Answer:

Activity 1: Since the only forces apposing the upward movement of the load are gravity and air resistance the formulas of which can both be solved with weight then with constant variables for gravity and air, effort IS directly proportional to weight.

Activity 2:

Inclined plane.

to make lifting easier; raising or lowering a load.

Explanation:

hope this helps.

4 0
3 years ago
Read 2 more answers
Determine the percent composition of sodium in sodium carbonate
crimeas [40]
11×2=22
(11×2)+(6)+(8×3)=52
22/52=0.4230
0.4230×100=42.3%
3 0
3 years ago
A student needs to prepare 100. mL of 0.612 M Cu(NO3)2 solution. What mass, in grams, of copper(II) nitrate should the student u
Temka [501]

Answer: 11.5 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution

Molarity=\frac{n\times 1000}{V_s}

where,

Morality = 0.612 M

n= moles of solute  

V_s = volume of solution in ml = 100 ml

Now put all the given values in the formula of molarity, we get

0.612=\frac{n\times 1000}{100ml}

n=0.0612moles

Mass={\text {moles of solute }}{\times {\text {molar mass}}=0.0612moles\times 187.56g/mol=11.5g

Therefore, the mass of copper (II)nitrate required is 11.5 grams

3 0
3 years ago
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