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3 years ago

What is formed when hydrobromic acid, hbr, and calcium hydroxide, ca(oh)2, are combined?

1 answer:

6 0

This is an acid – base reaction and this always result a salt and water in a neutralization reaction. 

The salt that is formed will be calcium bromide (calcium is located in group 2 so calcium bromide has a formula of CaBr2)

so essentially we got:

HBr + Ca(OH)2 ------> CaBr2 + H2O

balancing the elements: 

2HBr(aq) + Ca(OH)2(aq) --------> CaBr2(aq) + 2H2O(l)

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Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of alumin

Lerok [7]

Answer: The percentage yield of HF is 73.36 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For calcium fluoride:

Given mass of calcium fluoride = 6.25 kg = 6250 g (Conversion factor: 1 kg = 1000 g)

Molar mass of calcium fluoride = 78.07 g/mol

Putting values in above equation, we get:

$\text{Moles of calcium fluoride}=\frac{6250g}{78.07g/mol}=80.05mol$

For the given chemical reaction:

$CaF_2+H_2SO_4\rightarrow CaSO_4+2HF$

By Stoichiometry of the reaction:

1 mole of calcium fluoride produces 2 moles of hydrofluoric acid

So, 80.05 moles of calcium fluoride will produce = $\frac{2}{1}\times 80.05=160.1mol$ of hydrofluoric acid

Now, calculating the theoretical yield of hydrofluoric acid using equation 1, we get:

Moles of of hydrofluoric acid = 160.1 moles

Molar mass of hydrofluoric acid = 20.01 g/mol

Putting values in equation 1, we get:

$160.1mol=\frac{\text{Theoretical yield of hydrofluoric acid}}{20.01g/mol}=3203.6g=3.20kg$

To calculate the percentage yield of hydrofluoric acid, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of hydrofluoric acid = 2.35 kg

Theoretical yield of hydrofluoric acid = 3.20 kg

Putting values in above equation, we get:

$\%\text{ yield of hydrofluoric acid}=\frac{2.35g}{3.20g}\times 100\\\\\% \text{yield of hydrofluoric acid}=73.36\%$

Hence, the percentage yield of HF is 73.36 %

4 0

3 years ago

Which are the hybrid orbitals of the carbon atoms in the following molecules? (a) H3C―CH3 sp sp2 sp3 (b) H3C―CH═CH2 sp sp2 sp3 (

kherson [118]

Answer:

(a) sp³ sp³

H₃C - CH₃

(b) sp³ sp²

H₃C - CH = CH₂

sp²

H₃C - C ≡ C - CH₂OH

sp sp³

(d) sp³ sp²

H₃C - CH=O

Explanation:

Alkanes or the carbons with all the single bonds are sp³ hybridized.

Alkenes or the carbons with double bond(s) are sp² hybridized.

Alkynes or the carbons with triple bond are sp hybridized.

Considering:

(a) H₃C-CH₃ , Both the carbons are bonded by single bond so both the carbons are sp³ hybridized.

Hence,

sp³ sp³

H₃C - CH₃

(b) H₃C-CH=CH₂ , The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp² hybridized because they are bonded by double bond.

Hence,

sp³ sp²

H₃C - CH = CH₂

sp²

(c) H₃C-C≡C-CH₂OH , The carbons of the methyl group and alcoholic group are sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp hybridized because they are bonded by triple bond.

Hence,

sp³ sp

H₃C - C ≡ C - CH₂OH

sp sp³

(d)CH₃CH=O, The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The other carbon is sp² hybridized because it is bonded by double bond to oxygen.

Hence,

sp³ sp²

H₃C - CH=O

8 0

3 years ago

Listed in the Item Bank are key terms and expressions, each of which is associated with one of the columns. Some terms may displ

Elena-2011 [213]

I don't think it is possible.

4 0

3 years ago

Read 2 more answers

An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0025 M O2. At this temperature, Kc

pantera1 [17]

Answer:

$5.35 *10^{-4}$ M

Explanation:

Equation for the reaction is as follows:

$2CO_{(g)}$ + $O_{2(g)}$ ⇄ $2CO_{2(g)}$

By Applying the ICE Table; we have

$2CO_{(g)}$ + $O_{2(g)}$ ⇄ $2CO_{2(g)}$

Initial x 0.0025 M 0.0010 M

Change 0 0 0

Equilibrium x 0.0025 M 0.0010 M

$K_c =\frac{[CO_2]^2}{[CO]^2[O_2]}$

Given that $K_c = 1.4*10^2$ ; Then:

$1.4 *10^2 = \frac{(0.001)^2}{(x)^2(0.025)}$

$1.4 *10^2*0.025 = \frac{(0.001)^2}{(x)^2}$

$3.5 =( \frac{(0.001)}{(x)})^2$

$\sqrt {3.5} = \sqrt {( \frac{(0.001)}{(x)} )^2}$

$1.87=\frac{(0.001)}{(x)}$

$(x)= \frac{(0.001)}{1.87 }$

$x = 5.35 *10^{-4}$ M

∴ The equilibrium concentration of CO = $x = 5.35 *10^{-4}$ M

7 0

3 years ago

All the following are compounds except. A.water. B.sugar solution. C.aqueous. D.sulphur dioxide

kap26 [50]

Answer:

B.sugar solution

Explanation:

A sugar solution is not a compound.

Rather, the solution is a mixture and an impure substance.

A compound is composed of two or more kinds of atoms or elements joined in a definite grouping .
The properties of the compound is distinct from those of the individual atoms that makes them.
Sugar solution is sugar and water
It is made up of two compounds not chemically combined.

5 0

3 years ago