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SOVA2 [1]
4 years ago
13

In a titration 5.0 mL of a 2.0 M NaOH aq solution exactly neutralizes 10.0 of an HCL aq solution what is the concentration of th

e HCL(aq)solution
Chemistry
1 answer:
KonstantinChe [14]4 years ago
8 0
The  concentration    of  HCl solution  is  calculated  as  follows

write  the  equation   for  reaction

NaOH  + HCl = NaCl  +  H2O

find  the  moles   NaOH  =  molarity  x volume /1000

=  5  x2/1000  =  0.01  moles

by  use  of mole  ratio  between  NaOh  to HCl  which  is 1 :1  the  the  moles of  Hcl   is  also =  0.01 moles

concentration is therefore =   moles/volume    x1000

=  0.01/10  x1000 = 1M

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How can you increase the energy of a glass of water?
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3 years ago
Who is considered the founder of the periodic table of the elements?
nata0808 [166]

Answer:

Dmitry Mendeleev

Explanation:

Around 1869 a Russian scientist, Dmitry Mendeleev formed what is now known as the periodic table or chart. The Mendeleevian periodic table was based on the atomic weights of elements using the periodic law. The periodic law states that "chemical properties of elements are a periodic function of their atomic weights".

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7 0
3 years ago
URGENT CHEMISTRY EXPERT!
vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl

(b) Moles of CuCl₂

\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} =  \text{3.340 mmol CuCl}_{2}

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

\text{Moles of Na$_{3}$PO}_{4} =  \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}

(d) Volume of Na₃PO₄

V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}

Part 2. Net ionic equation

(a) Molecular equation

\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)  

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>  

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0
3 years ago
A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo
Liula [17]

<u>Answer:</u> The molar mass of the insulin is 6087.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g   (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = 62.364\text{ L.mmHg }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol

Hence, the molar mass of the insulin is 6087.2 g/mol

8 0
3 years ago
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