All categories

3 years ago

How does a lithium cation compare to a lithium atom

Chemistry

2 answers:

gayaneshka [121]3 years ago

8 0

C. The cation is smaller

Trava [24]3 years ago

7 0

How does a lithium cation compare to a lithium atom?C. the cation is smaller

You might be interested in

Equilibrium describes the balance that results from a forward and reverse reaction proceeding at equal rates. Since reaction rat

Delvig [45]

Answer:

b) the reaction proceeds to a new equilibrium in the direction that offsets the change.

Explanation:

According to Le Chatelier's principle, when a system experiences a constraint such as a change in pressure, temperature or concentration, the system will readjust itself in order to annul the constraint.

This simply means that when temperature, concentration or pressure is changed, a new equilibrium position is reached in order to offset the changes in the system.

5 0

3 years ago

At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank

hodyreva [135]

Answer:

$\large \boxed{\text{-92 $^{\circ}$C}}$

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

$\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}$

2. Temperature in degrees Celsius

$\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}$

8 0

3 years ago

A student measures the mass of a 6.0 cm3 block of brown sugar to be 10.0 g. What is the density of the brown sugar?

Alex17521 [72]

Answer:

1.67g/cm3

Explanation:

The formula for density is $d=\frac{m}{v}$ . The m variable stands for mass and the v variable stands for volume.

The mass of the brown sugar is 10.0g and the volume is 6.0cm3, so we can plug those values into the equation.

$d=\frac{10g}{6cm^{3} }$

$d=1.67$ $\frac{g}{cm^{3} }$

Rounded to 3 significant figures, the density of the block of brown sugar is 1.67 g/cm3. If the mass is in grams and the volume is in cm3, the unit for the final answer is $\frac{g}{cm^{3} }$ (grams per centimetres cubed).

8 0

4 years ago

Read 2 more answers

ang mga kumpanya ng electric motorcycle ay nag anunsyon na hindi na sila gagawa pa ng motor na pinaandar ng kuryente .ano sa pal

andreev551 [17]

:)) :)) :)) :)) :)) :)) :))

5 0

3 years ago

When .080 moles of propane burn at STP, what volume of carbon dioxide is produced?

ahrayia [7]

Taking into account the reaction stoichiometry and the definition of STP, 5.4 L of carbon dioxide is produced.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
H₂O: 4 moles

<h3>Moles of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of C₃H₈ form 3 moles of CO₂, 0.080 moles of C₃H₈ form how many moles of CO₂?

$amount of moles of CO_{2} =\frac{0.080 moles of C_{3} H_{8}x3 moles of CO_{2} }{1 mole of C_{3} H_{8} }$

<u><em>amount of moles of CO₂= 0.24 moles</em></u>

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Volume of CO₂ produced</h3>

You can apply the following rule of three: If by definition of STP 1 mole of CO₂ occupies 22.4 L, 0.24 moles of CO₂ how much volume does it occupy?

$volume of CO_{2} =\frac{0.24 molesx22.4 L}{1 mole}$