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2 years ago

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2 answers:

skad [1K]2 years ago

7 0

Answer:

c

c

Explanation:

mark me brainliest pls

Naya [18.7K]2 years ago

5 0

Coco beans poops okkkkk

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Gelneren [198K]

A solution <span>has a uniform composition and is only able to be separated by chemical means.</span>

4 0

2 years ago

Use the Henderson-Hasselbalch equation, eq. (3), to calculate the pH expected for a buffer solution prepared from this acid and

notka56 [123]

Answer:

$pH=4.56$

Explanation:

Hello there!

In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

$n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol$

And the concentrations are:

$[acid]=0.000855mol/0.025L=0.0342M$

$[base]=0.000781mol/0.025L=0.0312M$

Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

$pH=-log(2.5x10^{-5})+log(\frac{0.0312M}{0.0342M} )\\\\pH=4.60-0.04\\\\pH=4.56$

Best regards!

6 0

2 years ago

What is formed when two or more elements are bonded together?

Nadusha1986 [10]

Answer:

a compound

Explanation:

a compound consists of two or more elements combined

6 0

2 years ago

Read 2 more answers

A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<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)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0

3 years ago

What is a common land form that is formed when chemical weathering, specifically carbonation, is taking place?

skad [1K]

Answer:

plateaus

Explanation:

6 0

2 years ago