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Svetllana [295]
3 years ago
10

Determine the percent composition of Lithium in Li3PO3 (lower case BOTH 3)

Chemistry
1 answer:
zepelin [54]3 years ago
3 0

Answer:

A. 20.866%

Explanation:

First, we calculate molar mass of Li3PO3

Molar mass of Li3PO3 = 6.9 (3) + 31 + 16(3)

= 20.7 + 31 + 48

= 99.7g/mol

To calculate the percent composition of Lithium, we say:

Mass of lithium in compound/mass of compound × 100

% composition = 20.7/99.7 × 100

= 0.208 × 100

= 20.8%.

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Is potassium fluoride a covalent bond or ionic bond
Ivahew [28]

It's an ionic bond! Potassium is a cation, or a metal with a positive charge, and fluoride is an anion, or a nonmetal with a negative charge.

A covalent bond is the bond between two nonmetals.

Hope this helped!

7 0
4 years ago
Read 2 more answers
The specific heat of aluminum is 0.897j/(g•°C). If a 22.6 g sample of aluminum is heated from 25°C to 250°C, then how much heat
Oxana [17]

Q=mcat

=(22.6)(.897)(250-25)

4,561.245 or 4.5 * 10^-3

3 0
3 years ago
How many grams of CO₂ can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C₄H₁₀(g) + 13 O₂
Vilka [71]

Answer:

485.76 g of CO₂ can be made by this combustion

Explanation:

Combustion reaction:

2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

If we only have the amount of butane, we assume the oxygen is the excess reagent.

Ratio is 2:8. Let's make a rule of three:

2 moles of butane can produce 8 moles of dioxide

Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂

We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g

5 0
3 years ago
4.41 g of propane gas (C3H8) is injected into a bomb calorimeter and ignited with excess oxygen, according to the reaction below
gayaneshka [121]

Answer :  The heat of the reaction is -221.6 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter

q_{rxn}=-q_{cal}

q_{cal}=c_{cal}\times \Delta T

where,

q_{rxn} = heat released by the reaction = ?

q_{cal} = heat absorbed by the calorimeter

c_{cal} = specific heat of calorimeter = 97.1kJ/^oC=97100J/^oC

\Delta T = change in temperature = (T_{final}-T_{initial})=(27.282-25.000)=2.282^oC

Now put all the given values in the above formula, we get:

q_{cal}=(97100J/^oC)\times (2.282^oC)

q_{cal}=221582.2J=221.6kJ

As, q_{rxn}=-q_{cal}

So, q_{rxn}=-221.6kJ

Thus, the heat of the reaction is -221.6 kJ

6 0
3 years ago
Select the correct structure that
UkoKoshka [18]

Answer:

ugh this wont save anyway i think i'm just checking if ill save

Explanation:

if i does C both

hope this helps

6 0
3 years ago
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