Please help is it balanced or not and show me how you got it please I need this fast.

2. Determine the molarity of the NaOH solution in each trial. a. Trial 1 Molarity: b. Trial 2 Molarity: 3. Calculate the average

butalik [34]

Answer:

This question is incomplete

Explanation:

This question is incomplete but...

1) You can calculate the molarity of the NaOH for each trial by following the steps below.

The formula for Molarity (M) is

M = number of moles (n) ÷ volume (V)

where the unit of volume must be in Litres or dm³

The unit of molarity is mol/dm³ or mol/L or molar conc (M)

The final answer must have the unit of molarity

If the number of moles is not provided, look out for the mass of NaOH used and then calculate your number of moles (n) as

n = mass of NaOH used ÷ molar mass of NaOH

Where the atomic mass of sodium (Na) is 23, oxygen (O) is 16 and hydrogen (H) is 1. Hence, molar mass for NaOH is 23 + 16 + 1 = 40 g/mol

n = mass of NaOH used ÷ 40

2) Average Molarity will be (Trial 1 Molarity +Trial 2 Molarity) ÷ 2

Answer must be in mol/dm³ or mol/L or M

3) Label the volumentric flask containing the NaOH solution with the answer gotten from (2) above

3 0

3 years ago

PLZZZ HELP FASSSSSSSSSSSSTTTTTTTTTTTTT

aleksandrvk [35]

Answer:

i think multipling 30 to the 40

30 × 40 = 1200ppt

7 0

3 years ago

2. Iron has a density of 7.86 g/cm). A sample of an unknown metal is measured and found to be

alekssr [168]

The volume of sample = 2.25(1.5)(2.25) = 7.59375 cm3
Density = mass / volume
Density = 55.75 / 7.59375 = 7.34156 g/cm3

As the density of iron is 7.86 g/cm3 instead, it is not iron

3 0

3 years ago

If 20. 3 grams of an aqueous solution of iron(ii) chloride, fecl2, contains 2. 71 grams of iron(ii) chloride, what is the percen

Hatshy [7]

If 20. 3 grams of an aqueous solution of iron(ii) chloride contains 2. 71 grams of iron(ii) chloride, then percentage by mass of iron(ii) chloride in the solution is 13.34%

Calculation,

Given:

mass of an aqueous solution of iron(ii) chloride = 20. 3 grams

mass of iron(ii) chloride = 2. 71 grams

The percentage by Mass is equal to percentage of mass of solute present in total mass of solution. It is also known as Mass percentage.

percentage by Mass = 100× 2. 71 grams/20. 3 grams = 13.34%

Hence, the percentage by mass of iron(ii) chloride in the solution is 13.34% .

to learn more about mass percentage

brainly.com/question/27108588

#SPJ4

7 0

2 years ago

Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar

kakasveta [241]

Answer:

The answer is " $\bold{0.525\ \ atm^{-1}}$ "

Explanation:

Given equation:

$2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$

Given value:

$\Delta rH =-198.2 \ \ \frac{KJ}{mol}$

$Kp=1100 \ K$

$\Delta x = 2-(2+1)\\\\$

$= 2-(2+1)\\\\= 2-(3)\\\\= -1$

$\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right$

calculating the total pressure on equilibrium= $(1-2x)+(0.5-x)+2x \ atm\\\\$

$= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\$

$\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15\\\\$

calculating the pressure in $So_2$ :

$= (1-2 \times 0.15)$

$= 1-0.30 \\\\ =0.70 \ atm$

calculating the pressure in $O_2$ :

$= (0.5- 0.15)\\\\= 0.35 \ atm \\$

calculating the pressure in $So_3$ :

$= (2 \times 0.15)\\\\= (.30) \ atm \\\\$

Calculating the Kp at 1100 K:

$= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1} \\\\$

4 0

4 years ago