The more particles a substance has at a given temperature, the more thermal energy is has. True or false?

Use the equation editor or "Insert Chemistry - WIRIS editor" to write the balanced molecular chemical equation for the reaction

11Alexandr11 [23.1K]

Answer:

<u>Balanced equation:</u>

$Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)$

Explanation:

The chemical reaction between Lead(II) Nitrate and potassium carbonate is as follows.

$Lead(II)Nitrate+Potassium\,carbonate \rightarrow Lead(III)\,\,carbonate+Potassium\,nitrate$

$Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)$

<u>Ionic equation:</u>

$Pb^{2+}(aq)+2NO_{3}^{-}(aq)+2K^{+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)+K^{+}(aq)+2NO_{3}^{-}$

Cancel the same ions on the both sides of the reaction.

The net ionic equation is as follows.

$Pb^{2+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)$

4 0

3 years ago

in the reaction mgcl2 + 2koh mg(oh)2 + 2kcl, if 3 moles mgcl2 are added to 4 moles koh what determines how much mg(oh)2 is made

otez555 [7]

Every mole of MgCl2 reacts with 2 moles of KOH, therefore the 4 moles of KOH will only react with 2 moles of MgCl2, making it the limiting reagent and therefore KOH determines how much Mg(OH)2 is produced.

6 0

3 years ago

Read 2 more answers

What is the osmotic pressure, in torr, of a 3.00% solution of NaCl in water when the temperature of the solution is 45 ºC? Enter

Anit [1.1K]

213034 torr is the osmotic pressure.

Explanation:

osmotic pressure is calculated by the formula:

osmotic pressure= iCrT

where i= no. of solute

c= concentration in mol/litre

R= Universal Gas constant

T = temp

It is given that solution is 3% which is 3gms in 100 ml.

let us calculate the concentration in moles/litre

3gm/100ml*1000ml/1L*1mol NaCl/55.84g NaCl

= 5.372 gm/litre

Putting the values in the formula, Temp in Kelvin 318.5K

osmotic pressure= 2*5.372*0.083 * 318.5 Gas constant 0.083

= 284.023 bar or 213018 torr. c= 5.372 moles/L

i=2 for NaCl

4 0

3 years ago

16. The concentration of a solution of potassium hydroxide is determined by titration with nitric

____ [38]

Answer:

$M_{base}=0.709M$

Explanation:

Hello,

In this case, since the reaction between potassium hydroxide and nitric acid is:

$KOH+HNO_3\rightarrow KNO_3+H_2O$

We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:

$n_{acid}=n_{base}$

That in terms of molarities and volumes is:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, solving the molarity of the base (KOH), we obtain:

$M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} =\frac{0.498M*42.7mL}{30.0mL}\\ \\M_{base}=0.709M$

Regards.

3 0

3 years ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago