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2 years ago

(42.7g +0.259g) / (28.444mL x 12.367)

1 answer:

4 0

As you mentioned the question in chemistry so it may be a concentration unit I.e mass per volume.
(SOLUTION)
Mass/volume = total mass/ total volume
Now:
Total mass= 42.95 g
Total volume =351.8mL
Put these values in above equation;
Mass/volume = 42.95/351
“”. = 0.122 g/mL

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A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na

s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f: after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

$Normality=molarity \times n-factor$

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point: $N_1V_1=N_2V_2$

Before equivalence point : $N_1V_1>N_2V_2$

After the equivalence point: $N_1V_1$

$N_1V_1=100\times1=100$

case a: 5.00 mL of 1.00 M NaOH

$N_2V_2=5\times1=5$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

$N_2V_2=100\times1=100$

$N_1V_1=N_2V_2$ hence it is equivalence point

case c: 10.0 mL of 1.00 M NaOH

$N_2V_2=10\times1=10$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

$N_2V_2=150\times1=150$

$N_1V_1$ hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

$N_2V_2=50\times1=50$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

$N_2V_2=200\times1=200$

$N_1V_1$ hence it is after the eqivalence point

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3 years ago

Which methods could be used to dilute a solution of sodium chloride (NaCl)? Select all that apply.

Pavel [41]

Answer:

NaOH₂2CO3

Explanation:yes i know it is not a Balinese equation

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3 years ago

72.0 grams of water how many miles of sodium with react with it?

Flura [38]

Answer:

$\large \boxed{\text{8.00 mol}}$

Explanation:

We will need a balanced chemical equation with masses, moles, and molar masses.

1. Gather all the information in one place:

Mᵣ: 18.02

2Na + H₂O ⟶ 2NaOH + H₂

m/g: 72.0

2. Moles of H₂O

$\text{Moles of H$_{2}$O} = \text{72.0 g H$_{2}$O} \times \dfrac{\text{1 mol H$_{2}$O}}{\text{18.02 g H$_{2}$O}} = \text{3.996 mol H$_{2}$O}$

3. Moles of Na

The molar ratio is 2 mol Na/1 mol H₂O.

$\text{Moles of Na} = \text{3.996 mol H$_{2}$O} \times \dfrac{\text{2 mol Na}}{\text{1 mol H$_{2}$O}} = \textbf{8.00 mol Na}\\\\\text{The water will react with $\large \boxed{\textbf{ 8.00 mol}}$ of Na}$

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3 years ago

When the molecules of gases are heated, the average kinetic energy of the molecules increases,

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True is the answer to this

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3 years ago

I’m really bad at science and i need to know how to define it

postnew [5]

It is a column of elements in the periodic table of the chemical elements.

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3 years ago