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3 years ago

Convert mass to moles for both reactants

2 answers:

6 0

To convert mass to moles you can use the formula:

m = n*MM

OR

n = m/MM

Where n = number of moles
m = mass
MM = molar mass

Stolb23 [73]3 years ago

3 0

Do you have a picture of the problem?

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2. Lysosomes operate within the cell bye to the

Marat540 [252]

Answer:

Explanation:

Lysosomes contain digestive enzymes which are used for digestion of food and dead organelles. It fuses with a vacuole which contain the targeted material(food or dead organelles) and digests the material inside it

3 0

4 years ago

In a titration, 4.7 g of an acid (HX) requires 32.6 mL of 0.54 M NaOH(aq) for complete reaction. What is the molar mass of the a

katrin2010 [14]

Answer : The molar mass of an acid is 266.985 g/mole

Explanation : Given,

Mass of an acid (HX) = 4.7 g

Volume of NaOH = 32.6 ml = 0.0326 L

Molarity of NaOH = 0.54 M = 0.54 mole/L

First we have to calculate the moles of NaOH.

$\text{Moles of }NaOH=\text{Molarity of }NaOH\times \text{Volume of solution}=0.54mole/L\times 0.0326L=0.017604mole$

Now we have to calculate the moles of an acid.

In the titration, the moles of an acid will be equal to the moles of NaOH.

Moles of an acid = Moles of NaOH = 0.017604 mole

Now we have to calculate the molar mass of and acid.

$\text{Moles of an acid}=\frac{\text{Mass of an acid}}{\text{Molar mass of an acid}}$

Now put all the given values in this formula, we get:

$0.017604mole=\frac{4.7g}{\text{Molar mass of an acid}}$

$\text{Molar mass of an acid}=266.985g/mole$

Therefore, the molar mass of an acid is 266.985 g/mole

3 0

3 years ago

Which equation describes a reduction?

ira [324]

The answer choice is going to be B.

7 0

4 years ago

The concentration of a biomolecule inside a rod‑shaped prokaryotic cell is 0.0035 M . Calculate the number of molecules inside t

timama [110]

Answer: This rod-shape prokaryotic cell has 3.740,734725‬ molecules

Explanation:

Step 1 : given data

Molarity of the prokaryotic cell = 0.0035 M

Length of the cell = 4.2 μm = 4.2 * 10^-6 m

diameter of the cell = 1.3 μm = 1.3 * 10^-6 m

Step 2: calculate volume

To calculate volume of a rod, weneef to know the radius.

V = r ² × l

The radius = half of the diameter : r = d/2 ⇒ (1.3 * 10^-6 m)/2 = 0.65 * 10^-6 m

V= (0.65 * 10^-6 m)² * 4.2 * 10^-6 m = 1.7745 * 10 ^-18 L

Step 3: Calculating number of moles

Number of moles = Concentration * Volume

moles = 0.0035 M * 1.7745 * 10 ^-18 L = 6.21075 * 10^-21 moles‬

Step 4: calculating number of molecules

1 mole contains 6.023 * 10 ^-23 molecules

6.21075 * 10^-21 moles contain : 6.21075 * 10^-21 * 6.023 * 10 ^-23 molecules = 3.740,734725‬ molecules

This rod-shape prokaryotic cell has 3.740,734725‬ molecules

4 0

3 years ago

Acetic acid has a pKa of 4.74. Buffer A: 0.10 M HC2H3O2, 0.10 M NaC2H3O2 Buffer B: 0.30 M HC2H3O2, 0.30 M NaC2H3O2 Buffer C: 0.5

e-lub [12.9K]

Answer:

Buffer B has the highest buffer capacity.

Buffer C has the lowest buffer capacity.

Explanation:

An effective weak acid-conjugate base buffer should have pH equal to $pK_{a}$ of the weak acid. For buffers with the same pH, higher the concentrations of the components in a buffer, higher will the buffer capacity.

Acetic acid is a weak acid and $CH_{3}COO^{-}$ is the conjugate base So, all the given buffers are weak acid-conjugate base buffers. The pH of these buffers are expressed as (Henderson-Hasselbalch):

$pH=pK_{a}(CH_{3}COOH)+log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}$

$pK_{a}(CH_{3}COOH)=4.74$

Buffer A: $pH=4.74+log(\frac{0.10}{0.10})=4.74$

Buffer B: $pH=4.74+log(\frac{0.30}{0.30})=4.74$

Buffer C: $pH=4.74+log(\frac{0.10}{0.50})=4.04$

So, both buffer A and buffer B has same pH value which is also equal to $pK_{a}$ . Buffer B has higher concentrations of the components as compared to buffer A, Hence, buffer B has the highest buffer capacity.

The pH of buffer C is far away from $pK_{a}$ . Therefore, buffer C has the lowest buffer capacity.

6 0

3 years ago