What are the examples of Alkaline Earth metals?

How would you prepare 250 mL of a 0.100 M solution of fluoride ions from solid CaF2?

andrey2020 [161]

In 250 mL of volumetric flask add 0.975875 grams of $CaF_2$ and dissolve it in the 250 mL of water.

Given:

The solid of calcium fluoride.

To prepare:

The 250 mL solution of 0.100 M of fluoride ions from solid calcium fluoride.

Method:

Molarity of the fluoride ion solution needed = M = 0.100 M

The volume of the fluoride ion solution needed = V = 250 mL

$1 mL = 0.001L\\V=250 mL=250\times 0.001 L=0.250 L$

The moles of fluoride ion needed = n

According to the definition of molarity:

$M=\frac{n}{V}\\0.100M=\frac{n}{0.250 L}\\n=0.100M\times 0.250 L=0.025 mol$

Moles of fluoride ion = 0.025 mol

We know that solid calcium fluoride dissolves in water to give calcium ions and fluoride ions.

$CaF_2(s)\rightarrow Ca^{2+}(aq)+2F^-(aq)$

According to reaction, 2 moles of fluoride ions are obtained from 1 mole of calcium fluoride, then 0.025 moles of fluoride ions will be obtained from:

$=\frac{1}{2}\times 0.025 mol=0.0125 \text{mol of } CaF_2$

Moles of calcium fluoride = 0.0125 mol

Mass of calcium fluoride needed to prepare the solution :

$=0.0125 mol\times 78.07 g/mol=0.975875 g$

Preparation:

Weight 0.975875 grams of calcium fluoride
Add weighed calcium fluoride to a volumetric flask of the labeled volume of 250 mL.
Now add a small amount of water to dissolve the calcium fluoride completely.
After this add more water up to the mark of the volumetric flask of volume 250 mL.

Learn more about molarity of solution ere:

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3 0

2 years ago

A gas exerts a pressure of 1.8 atm at a temperature of 60 degrees celsius. What is the new temperature when the pressure of the

Kamila [148]

The answer for the following problem is mentioned below.

Therefore the final temperature of the gas is 740 K

Explanation:

Given:

Initial pressure of the gas ( $P_{1}$ ) = 1.8 atm

Final pressure of the gas ( $P_{2}$ ) = 4 atm

Initial temperature of the gas ( $T_{1}$ ) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas ( $T_{2}$ )

We know;

From the ideal gas equation;

we know;

P × V = n × R × T

So;

we can tell from the above equation;

P ∝ T

(i.e.)

$\frac{P}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

$P_{1}$ = initial pressure of a gas

$P_{2}$ = final pressure of a gas

$T_{1}$ = initial temperature of a gas

$T_{2}$ = final temperature of a gas

$\frac{1.8}{4}$ = $\frac{333}{T_{2} }$

$T_{2}$ = $\frac{333*4}{1.8}$

$T_{2}$ = 740 K

Therefore the final temperature of the gas is 740 K

8 0

2 years ago

Aspartame is a sweetner that is 160 times sweeter than sucrose (table sugar) when dissolved in water. The molecular formula of a

NikAS [45]

Answer:

294.307 g/mol

Explanation:

The first question for this statment is:

Calculate the gram-formula-mass of aspartame.

<h2>Solution</h2>

The chemical formula is:

$C_{14}H_{18}N_2O_5 .$

The gram-formula-mass is calculated adding the masses for all the atoms in the molecular formula:

Atom Number of atoms Atomic mass Total mass

g/mol g/mol

C 14 12.011 14 × 12.011 = 164.154

H 18 1.008 18 × 1.008 = 18.144

N 2 14.007 2 × 14.007 = 28.014

O 5 15.999 5 × 15.999 = 79.995

===================

Total 294.307 g/mol

Answer: 294.307 g/mol

6 0

2 years ago

Newton’s second law of motion is F=ma A net force of 60 N north acts on an object with a of 30 kg. Use Newton’s second law of mo

horsena [70]

I’m pretty sure the immediate answer is 2. Using the equation 60N=(30kg)x2(whatever unit of measure) also it will change because if you change the mass or the Net Force, your going to have to redo the equation based on the new information.

8 0

3 years ago

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Rain showers have continued over a particular area for several days which of these weather factors is most likely responsible A.

Ede4ka [16]

D. a warm front causes rain showers to occur over a particular area for several days

7 0

2 years ago

Read 2 more answers