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3 years ago

How many moles are in 15 grams of lithium?

Chemistry

1 answer:

stira [4]3 years ago

8 0

Moles of substance = mass of that substance/molar mass of that substance = 15/6.9 = 2.17 moles.

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List of charge of elements that do not form compound easily

adelina 88 [10]

Answer:

The highlighted words in the explanation.

Explanation:

A clue comes by considering the noble gas elements, the rightmost column of the periodic table. These elements—helium, neon, argon, krypton, xenon, and radon—do not form compounds very easily, which suggests that they are especially stable as lone atoms. What else do the noble gas elements have in common?

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3 years ago

What is matter?

belka [17]

A Anything that has mass and takes up space

4 0

3 years ago

Read 2 more answers

What volume will 0.405 g of krypton gas occupy at STP?

Rufina [12.5K]

Answer:

The answer to your question is V = 0.108 L or 108 ml

Explanation:

Data

Volume = ?

mass = 0.405 g

Temperature = 273°K

Pressure = 1 atm

Process

1.- Convert mass of Kr to moles

83.8 g of Kr -------------------- 1 mol

0.405 g ------------------- x

x = (0.405 x 1) / 83.8

x = 0.0048 moles

2.- Use the Ideal gas law to solve this problem

PV = nRT

- Solve for V

V = nRT / P

- Substitution

V = (0.0048)(0.082)(273) / 1

- Simplification

V = 0.108 / 1

- Result

V = 0.108 L

8 0

3 years ago

Read 2 more answers

Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to the following equati

Scrat [10]

Here's your answer, hope this helps!

6 0

3 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

7 0

2 years ago