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3 years ago

Which quantity is equivalent to 39 grams of LiF?

Chemistry

2 answers:

Nutka1998 [239]3 years ago

3 0

I think that it is 1.5 mole it might not be

stich3 [128]3 years ago

3 0

Answer:

This question is incomplete, the completed question is below.

Which quantity is equivalent to 39 grams of LiF?

(a) 1.0 mole

(b) 2.0 mole

(d) 1.5 mole

The correct option is d

Explanation:

LiF is Lithium flouride. Lithium (Li) has a molar mass of 7 g/mol while flourine (F) has a molar mass of 19 g/mol. Hence, 1 mole of LiF will be

7 + 19 = 26 g/mol

If 1 mole of LiF = 26g,

X mole of LiF = 39g

When you cross multiply, you will have

X mole x 26g = 1 mole x 39g

make X the subject and you will have

X mole = 1 mole x 39g/26g

X mole = 1.5 mole

Hence, 1.5 mole of LiF is equivalent to 39 grams of LiF

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What is the empirical formula for the following molecular formula: C10H5O2

Tju [1.3M]

The empirical formula is the same as the molecular formula : C₁₀H₅O₂

<h3>Further explanation</h3>

Given

Molecular formula : C₁₀H₅O₂

Required

The empirical formula

Solution

The empirical formula (EF) is the smallest comparison of atoms of compound forming elements.

The molecular formula (MF) is a formula that shows the number of atomic elements that make up a compound.

(empirical formula) n = molecular formula

(EF)n=MF

(EF)n = C₁₀H₅O₂

If we divide by the number of moles of Oxygen (the smallest) which is 2 then the moles of Hydrogen will be a decimal number (not whole), which is 2.5, then the empirical formula is the same as the molecular formula

4 0

3 years ago

Enter your answer in the provided box. Calcium hydroxide may be used to neutralize (completely react with) aqueous hydrochloric

NNADVOKAT [17]

Answer:

49.95 g of HCl

Explanation:

Let's formulate the chemical equation involved in the process:

Ca(OH)2 + 2 HCl → CaCl2 + 2 H2O

This means that we need 1 mole of Calcium hydroxide to neutralize 2 moles of hydrochloric acid. From this, we calculate the quantity of HCl moles that would be neutralized by 0.685 moles of Ca(OH)2

1 mole Ca(OH)2 ---- 2 moles HCl

0.685 moles Ca(OH)2 ---- x = 1.37 moles HCl

Now that we know the quantity of HCl moles that would react, let's calculate the quantity of grams this moles represent:

1 mole of HCl ---- 36.46094 g

1.37 moles ------ x = 49.95 g of HCl

3 0

3 years ago

Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib

KengaRu [80]

Answer:

For methanol: Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol: Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol: Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

Explanation:

Given:

Mass of Methanol = 1.0 g

Mass of ethanol = 1.00 g

Mass of n-propanol = 1.00 g

For methanol:

2 CH₃OH + 3 O₂ ----> 2 CO₂ + 4 H₂O, ∆H₀ = -22.6 kJ/g (negative sign signifies release of heat)

1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 32.04 g/mol

Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol:

C₂H₅OH + 3 O₂ ----> 2 CO₂ + 3 H₂O, ∆H₀ = -29.7 kJ/g (negative sign signifies release of heat)

1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol:

2 C₃H₇OH + 9 O₂ ----> 6 CO₂ + 8 H₂O, ∆H₀ = -33.4 kJ/g , (negative sign signifies release of heat)

1 g of methanol on combustion gives 33.4 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 60.09 g/mol

Thus moles of methanol = 1 g/ (60.09 g/mol) = 0.0166 moles

Hence energy in kJ/mol:

Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

5 0

3 years ago

Electrons move in what is called an atoms electron _______. Plz help me

kupik [55]

Electrons move in what is called an electron cloud. Each atom has its own electron cloud associated with it. This region simply represents an area or likelihood within that an electron of corresponding energy can be found within.

5 0

3 years ago

How many moles of k3po4 can be formed when 4.4 moles of h3po4 react with 3.8 moles of koh? h3po4 + koh yields h2o + k3po4 be sur

schepotkina [342]

the balanced chemical equation for the reaction is as follows

H₃PO₄ + 3KOH ---> K₃PO₄ + 3H₂O

stoichiometry of H₃PO₄ to KOH is 1:3

first we have to find which the limiiting reactant is

as the amount of product formed depends on the amount of limiting reactant present

number of H₃PO₄ moles reacted - 4.4 mol

if H₃PO₄ is the limiting reactant

1 mol of H₃PO₄ reacts with 3 mol of KOH

then 4.4 mol of H₃PO₄ reacts with - 3 x 4.4 mol = 13.2 mol of KOH

but only 3.8 mol of KOH is present

therefore KOH is the limiting reactant

stoichiometry of KOH to K₃PO₄ is 3:1

number of KOH moles reacted - 3.8 mol

therefore number of K₃PO₄ formed = number of KOH moles reacted / 3

= 3.8 mol / 3 = 1.3 mol

answer is 1.3 mol of K₃PO₄

3 0

3 years ago