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3 years ago

What is the molas mass of lithium oxide li2o

Chemistry

2 answers:

gladu [14]3 years ago

7 0

The molar mass of lithium oxide is
29.88 g/mol

Alexxandr [17]3 years ago

6 0

Explanation:

Atomic mass means the sum of total number of protons and neutrons present in an element.

Molar mass is defined as the mass of one mole of a compound present in grams.

For example, molar mass of $Li_{2}O$ is as follows.

Molar mass of $Li_{2}O$ = 2 × atomic mass of Li + atomic mass of O

= $2 \times 6.94 g/mol + 16.0 g/mol$

= 13.88 g/mol + 16.0 g/mol

= 29.88 g/mol

Thus, we can conclude that molar mass of $Li_{2}O$ is 29.88 g/mol.

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Answer:

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What does phenolphthalein turn pink?

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3 years ago

Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everything from fats and waxes to air

Yanka [14]

Answer: a. 79.6 s

b. 44.3 s

c. 191 s

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$t_{\frac{1}{2}}=\frac{0.693}{8.7\times 10^{-3}s^{-1}}=79.6s$

b) for completion of 32% of reaction

$t=\frac{2.303}{k}\log\frac{100}{100-32}$

$t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{68}$

$t=44.3s$

c) for completion of 81 % of reaction

$t=\frac{2.303}{k}\log\frac{100}{100-81}$

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4 0

3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

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3 years ago

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stealth61 [152]

Answer:

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Explanation:

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