Sodium (Na) A Homogeneous Mixture B Element C Compound D Heterogeneous Mixture

What is the concentration of 10.00 mL of HBr if it takes 16.73 mL of a 0.253 M LiOH solution to neutralize it?

Leni [432]

First. let's write the reaction formula: HBr +LiOH ----> LiBr + H₂O

let's get the moles of LiOH first

moles= Molarity x Liters

moles= 0.253 M x 0.01673 Liter= 0.00423 moles LiOH

using the balanced equation, you can see that 1 mol LiOH is equal to 1 mol HBr. so:

0.00423 mol LiOH = 0.00423 mol HBr

now let's find the concentration

molarity= mol/ Liters

0.00423 mol/ 0.01000 Liters= 0.423 M

3 0

3 years ago

Explain why and how theories may be changed or replaced over time.

malfutka [58]

There are new ideas and questions that people are trying out and solving. Based off of that, the theories and ideas that we have now will change and evolve with the theories that have been tested and the questions that have been answered.

8 0

2 years ago

Read 2 more answers

Please help i cant find any answers...

Hitman42 [59]

Answer:

98.8

Explanation:

CsF + XeF6 --> CsXeF7

37.8g ................. ?g

37.8g CsF x (1 mol CsF / 151.9g CsF) x (1 mol CsXeF7 / 1 mol CsF) x (397.2g CsXeF7 / 1 mol CsXeF7) = 98.8g CsXeF7 .......... to three significant digits

8 0

2 years ago

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A small hole in the wing of a space shuttle requires a 15.9 cm2 patch.

Rzqust [24]

The patch area in square kilometers is 1.59*10⁻⁹ km₂

Why?

This is an unit conversion problem. We have to convert from cm² to km². We can do that by knowing that there are 100 cm in 1 m, and 1000 m in 1 km, so 100000 cm=1km. Knowing that we can apply the following conversion factor:

$15.9 cm^{2} * (\frac{1 km}{100000 cm} )^{2}=0.00000000159 km^{2}$

Now to convert this value to scientific notation, we have to move the decimal point to the right until we get a whole number, and the exponent of the number 10 is going to be the number of spaces we moved to the right (negative), so the final answer is:

$0.00000000159 km^{2}=1.59*10^{-9} km^{2}$

Have a nice day!

#LearnwithBrainly

8 0

3 years ago

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

Lerok [7]

<u>Answer:</u> The percentage yield of HF is 73.36 %

<u>Explanation:</u>

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

2 years ago