Complete the equation for the dissociation of Na2CO3(aq).Omit water from the equation because it is understood to be present.

matrenka [14]

<span>it is located directly under the sima</span>

4 0

3 years ago

Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

<u>Calculate the moles of potassium hydroxide.</u>

$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0

2 years ago

Which type of radioactive decay produces

sladkih [1.3K]

Which type of radioactive decay produces particles with the most mass? Alpha.

6 0

3 years ago

8.0 mol AgNO3 reacts with 5.0 mol Zn in

Yakvenalex [24]

Taking into account the reaction stoichiometry, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgNO₃ + Zn → 2 Ag + Zn(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

AgNO₃: 2 moles
Zn: 1 mole
Ag: 2 moles
Zn(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Zn reacts with 2 moles of AgNO₃, 5 moles of Zn reacts with how many moles of AgNO₃?

$amount of moles of AgNO_{3}= \frac{5 moles of Znx2 moles of AgNO_{3}}{1 mole of Zn}$

<u><em>amount of moles of AgNO₃= 10 moles </em></u>

But 10 moles of AgNO₃ are not available, 8 moles are available. Since you have less moles than you need to react with 5 moles of Zn, AgNO₃ will be the limiting reagent.

<h3>Moles of Ag formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 2 moles of AgNO₃ form 2 moles of Ag, 8 moles of AgNO₃ form how many moles of Ag?

$amount of moles of Ag=\frac{8 moles of AgNO_{3}x2 moles of Ag }{2 moles of AgNO_{3}}$