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Nimfa-mama [501]
3 years ago
12

You need to make an aqueous solution of 0.215 M aluminum bromide for an experiment in lab, using a 300 mL volumetric flask. How

much solid aluminum bromide should you add
Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
3 0

Answer:

We have to add 17.2 grams of aluminium bromide

Explanation:

Step 1: Data given

Molarity of the aluminum bromide solution = 0.215 M

Volume = 300 mL = 0.300 L

Molar mass aluminium bromide = 266.69 g/mol

Step 2: Calculate moles Aluminium bromide

moles AlBr3 = volume * molarity

Moles AlBr3 = 0.300 L * 0.215 M

Moles AlBR3 = 0.0645 moles

Step 3: Calculate mass aluminium bromide

Mass aluminium bromide = moles AlBr3 * molar mass AlBr3

Mass aluminium bromide = 0.0645 moles * 266.69 g/mol

Mass aluminium bromide = 17.2 grams

We have to add 17.2 grams of aluminium bromide

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New questionWhat mass of calcium chloride (CaCl₂) would beproduced from the reaction of 125.9 g of hydrochloriacid (HCI) with ex
marshall27 [118]

Answer:

191.6 g of CaCl₂.

Explanation:

What is given?

Mass of HCl = 125.9 g.

Molar mass of CaCl₂ = 110.8 g/mol.

Molar mass of HCl = 36.4 g/mol.

Step-by-step solution:

First, we have to state the chemical equation. Ca(OH)₂ react with HCl to produce CaCl₂:

Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O.

Now, let's convert 125.9 g of HCl to moles using the given molar mass (remember that the molar mass of a compound can be found using the periodic table). The conversion will look like this:

125.9\text{ g HCl}\cdot\frac{1\text{ mol HCl}}{36.4\text{ g HCl}}=3.459\text{ moles HCl.}

Let's find how many moles of CaCl₂ are being produced by 3.459 moles of HCl. You can see in the chemical equation that 2 moles of HCl reacted with excess Ca(OH)₂ produces 1 mol of CaCl₂, so we state a rule of three and the calculation is:

3.459\text{ moles HCl}\cdot\frac{1\text{ mol CaCl}_2}{2\text{ moles HCl}}=1.729\text{ moles CaCl}_2.

The final step is to find the mass of CaCl₂ using the molar mass of CaCl₂. This conversion will look like this:

1.729\text{ moles CaCl}_2\cdot\frac{110.8\text{ g CaCl}_2}{1\text{ mol CaCl}_2}=191.6\text{ g CaCl}_2.

The answer would be that we're producing a mass of 191.6 g of CaCl₂.

4 0
1 year ago
Which of the following happens when chlorine forms an ion?
Jlenok [28]

What  happens  when  chlorine  form  an ion  is that  it gains an electron and  has  an octet  in its  outer shell  ( answer  A)


<u><em> Explanation</em></u>

<u><em> </em></u>Chlorine is  is in atomic  number  17  in periodic table.

The electron configuration  of chlorine  is    1S2 2S2 2P6 3S2 3P5   or[Ne]3S2 3p5  or  2.8.7.

chlorine therefore  has 7 valence electron therefore it  gain  1 electron  to form Cl- ( ion)

Cl- has  8  electron in its outer  shell (  it  obeys  octet  rule  of eight valence in outer shell.

7 0
3 years ago
Read 2 more answers
You collect 552 mL of argon gas at 23.0 C. What volume will the gas occupy at 46.0C if the pressure remains constant?
taurus [48]

Answer:

1027.9 mL

Explanation:

Formula P1 x V1 / T1 = P2 x V2 / T2

Fill in what you know

Pressure is constant so no need to put that in making the formula

V1 / T1 = V2 / T2

Voulme 1= 950 mL

Volume 2= ?

Temperature 1 = 25 C

Temperature 2 = 50 C

Explanation:

Formula P1 x V1 / T1 = P2 x V2 / T2

Fill in what you know

Pressure is constant so no need to put that in making the formula

V1 / T1 = V2 / T2

Voulme 1= 950 mL

Volume 2= ?

Temperature 1 = 25 C

Temperature 2 = 50 C

8 0
2 years ago
What was formed as result of diverging plates
irina [24]

It causes earthquakes say you had to glass plates you eat from and smash them together and they make a loud sound and brake into each other thats like the diverging plates

6 0
3 years ago
Submit What is the solubility of Cd3(POA) 2 in water? (Ksp of Cd3(PO4)2 is 2.5 x 10-33) | 1 2 3 +/- . 0 x100
vesna_86 [32]

<u>Answer:</u> The solubility of Cd_3(PO_4)_2 in water is 1.18\times 10^{-7}mol/L

<u>Explanation:</u>

The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

Cd_3(PO_4)_2\rightleftharpoons 3Cd^{2+}+2PO_4^{3-}

                      3s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2

We are given:

K_{sp}=2.5\times 10^{-33}

Putting values in above equation, we get:

2.5\times 10^{-33}=(3s)^3\times (2s)^2\\\\2.5\times 10^{-33}=108s^5\\\\s=1.18\times 10^{-7}mol/L

Hence, the solubility of Cd_3(PO_4)_2 in water is 1.18\times 10^{-7}mol/L

6 0
3 years ago
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