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AnnZ [28]
3 years ago
10

Omg pls help i dunno what the frick frack this is

Chemistry
1 answer:
snow_lady [41]3 years ago
4 0

Answer:

1. Mass of KCl produced = 774.8 g of KCl

2. Mass of KNO₃ produced = 13.837g

3. Moles of NaOH made = 0.846 moles

4. Moles of LiCl produced = 0.846 moles

5. Moles of CO₂ produced = 207.6 moles

Explanation:

1. From the equation of reaction, 1 mole of ZnCl₂ produces, 2 moles of KCl.

5.02 moles of ZnCl₂ will produce, 2 × 5.02 moles of KCl = 10.4 moles of KCl

Molar mass of KCl = (39 + 35.5) g/mol = 74.5 g/mol

10.4 moles of KCl = 10.4 × 74.5 g

Mass of KCl produced = 774.8 g of KCl

2. Mole ratio of KNO₃ and KOH = 1:1

O.137 moles of KOH will produce 0.137 moles of KNO₃

Molar mass of KNO₃ = 101 g/mol

Mass of KNO₃ produced = 0.137 × 101 g = 13.837g

3. Molar mas of Ca(OH)₂ = 74.0 g

Moles of Ca(OH)₂ in 31.3 g = 31.3/74.0 = 0.423 moles of Ca(OH)₂

Mole ratio of NaOH and Ca(OH)₂ in the reaction = 2 : 1

Moles of NaOH made = 2 × 0.423 = 0.846 moles

4. Molar mass of MgCl₂ = 95.0 g

Moles of MgCl₂ in 40.2 g = 40.2/95.0 = 0.423 moles

From the reaction equation, mole ratio of MgCl₂ and LiCl = 1:2

Moles of LiCl produced = 2 × 0.423 = 0.846 moles

5. From the equation of reaction, 1 mole of C₆H₁₀O₅ produces 6 moles of cO₂

34.6 moles of C₆H₁₀O₅ will produce 34.6 × 6 moles of CO₂

Moles of CO₂ produced = 207.6 moles

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The procedure, which can be used to determine more accurately the concentration of the unknown acid is TO BACK-TITRATE WITH ADDITIONAL HYDROCHLORIC ACID TO NEUTRALIZE THE ADDITIONAL SODIUM HYDROXIDE THAT WAS ADDED.
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The back titration method is typically used when one needs to determine the concentration of an analyte provided there is a known molar concentration of excess reactants. 
From the information given in the question above, we are told that excess NaOH was added. To correct this mistake, the right thing to do is to use additional HCl to carry out back titration, taking note of the quantity of acid that will be needed to neutralize the excess NaOH.
7 0
3 years ago
For the following reaction, 5.05 grams of copper are mixed with excess silver nitrate. The reaction yields 11.0 grams of copper(
andreyandreev [35.5K]

Answer:

14.9 g is the ideal yield of Cu(NO₃)₂

Explanation:

Reactants for the reaction: Cu and AgNO₃

Products: Copper nitrate and Ag

The balanced reaction is: Cu(s) + 2AgNO₃(aq) →  2Ag (s) + Cu(NO₃)₂

As the silver nitrate is in excess, the Cu will be the limiting reagent.

We convert the mass to moles → 5.05 g . 1 mol/ 63.55 g = 0.0794 moles

Ratio is 1:1, so 0.0794 moles will produce 0.0794 moles of Cupper(II) nitrate. We convert the moles to mass, and that value will be the theoretical yield.

0.0794 mol . 187.55 g /1 mol =  of Cu(NO₃)₂

8 0
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Arrange the following elements based on their size (atomic radii) from largest to smallest: Ca, Ge, Br, K, Kr
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7 0
2 years ago
What mass of iron would be required to react with 10.0L of O2 gas?
SashulF [63]

33.2 g of iron (Fe)

Explanation:

We have the following chemical reaction in which iron react with oxygen to form the iron (III) oxide:

4 Fe + 3 O₂ → 2 Fe₂O₃

number of moles = volume / 22.4 (L/mole)

number of moles of O₂ = 10 / 22.4

number of moles of O₂ = 0.446 moles

Taking in account the chemical reaction we devise the following reasoning:

if           4 moles of Fe react with 3 moles of O₂

then     X moles of Fe react with 0.446 moles of O₂

X = (4 × 0.446) / 3 = 0.595 moles of Fe

number of moles = mass / molar weight

mass =  number of moles × molar weight

mass of Fe = 0.595 × 55.8

mass of Fe = 33.2 g

Learn more about:

number of moles

brainly.com/question/14167442

brainly.com/question/14122510

#learnwithBrainly

4 0
3 years ago
Ok so i haven't been on this in a long time ig and so i don't know i just need help on this so i can play my fav video game oof
Rama09 [41]

Answer:

J

Explanation:

its J

7 0
3 years ago
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