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A chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water. What is the concentration of the final solution? A

anygoal [31]

this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.

c1v1 = c2v2

where c1 is concentration and v1 is volume of the concentrated solution

and c2 is concentration and v2 is volume of the diluted solution to be prepared

50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL

substituting these values in the formula

1.50 M x 50.0 mL = C x 250 mL

C = 0.300 M

concentration of the final solution is A) 0.300 M

4 0

3 years ago

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5Br−+BrO3−+6H+→3Br2+3H2O

sashaice [31]

Explanation :

The balanced chemical reaction is,

$5Br^-+BrO_3^-+6H^+\rightarrow 3Br_2+3H_2O$

The expression for the rates of consumption of the reactants are:

The rate of consumption of $Br^-$ = $-\frac{1}{5}\frac{d[Br^-]}{dt}$

The rate of consumption of $BrO_3^-$ = $-\frac{d[BrO_3^-]}{dt}$

The rate of consumption of $H^+$ = $\frac{1}{6}\frac{d[H^+]}{dt}$

The expression for the rates of formation of the products are:

The rate of consumption of $Br_2$ = $+\frac{1}{3}\frac{d[Br_2]}{dt}$

The rate of consumption of $H_2O$ = $+\frac{1}{3}\frac{d[H_2O]}{dt}$

5 0

3 years ago

H2SO4 +HI → __ H2S+12 +H2O balance the equation

Citrus2011 [14]

Answer:

H2SO4 + 8HI → H2S + 4I2 + 4H2O

3 0

2 years ago

A gas mixture contains 10.0 mole% H2O (v) and 90.0 mole % N2. The gas temperature and absolute pressure at the start of each of

Rainbow [258]

Answer: (a). T = 38.2 °C (b). V = 1.3392 cm³ (c). ii and iii

Explanation:

this is quite easy to solve, i will give a step by step analysis to solving this problem.

(a). from the question we have that;

the Mole fraction of Nitrogen, yи₂ = 0.1

Also the Mole fraction of Water, yн₂o = 0.1

We know that the vapor pressure is equal to the partial pressure because the vapor tends to condense at due point.

ρн₂o = ṗн₂o

= yн₂oP = 0.1 × 500 mmHg = 50 mmHg

from using Antoine equation, we apply the equation

logρн₂o = A - B/C+T

T = B/(A - logρн₂o) - C

= 1730.63 / (8.07131 - log 50mmHg) - 223.426

T = 38.2 °C

We have that the temperature for the first drop of liquid form is 38.2 °C

(b). We have to calculate the total moles of gas mixture in a 30 litre flask;

n = PV/RT

n = [500(mmHG) × 30L] / [62.36(mmHGL/mol K) × 323.15K] = 0.744 mol

Moles of H₂O(v) is 0.1(0.744) = 0.0744 mol

Moles of N₂ is 0.9(0.744) = 0.6696 mol

we have that the moles of water condensed is 0.0744 mol i.e the water vapor in the flask is condensed

Vн₂o = 0.0744 × 18 / 1 (g/cm³)

Vн₂o = 1.3392 cm³

Therefore, the volume of the liquid water is 1.3392 cm³

(3). (ii) and (iii)

The absolute pressure of the gas and The partial pressure of water in the gas would change if the barometric pressure drops.

cheers i hope this helps!!!!

4 0

3 years ago

Before tackling this problem, be sure you know how to find the antilog of a number using a scientific calculator.

dybincka [34]

<h2>Question:- </h2>

A solution has a pH of 5.4, the determination of [H+].

<h2>Given :- </h2>

pH:- 5.4
pH = - log[H+]

<h2>To find :- concentration of H+</h2>

<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>

<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>

Take negative to other side

-pH = log H+

multiple Antilog on both side

(Antilog and log cancel each other )

Antilog (-pH) = [ H+ ]

New Formula :- Antilog (-pH) = [+H]

Now put the values of pH in new formula

Antilog (-5.4) = [+H]

we can write -5.4 as (-6+0.6) just to solve Antilog

Antilog ( -6+0.6 ) = [+H]

Antilog (-6) × Antilog (0.6) = [+H]

$Antilog (-6) = {10}^{ - 6} , \\ Antilog (0.6) = 4$

put the value in equation

${10}^{ - 6} \times 4 = [H+] \\ 4 \times {10}^{ - 6} = [H+]$

7 0

2 years ago

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