1. First, let's determine the amount of gas in each of the vessels after equilibrium is established

$$
\begin{aligned}
& \left\{\begin{array}{l}
v_{1}+v_{2}=4 \\
v_{2}-v_{1}=0.62
\end{array}\right. \\
& v_{2}=v_{1}+0.62 \\
& 2 v_{1}+0.62=4 \\
& v_{1}=1.69 \text { mol, } \\
& v_{2}=4-1.69=2.31 \text { mol. }
\end{aligned}
$$

Determining the amount of gas in each of the vessels - 2 points

In the equilibrium state, the pressures in both vessels equalize:

$$
p_{1}=p_{2}
$$

Therefore, according to the Clapeyron-Mendeleev equation

Then
Answer: $\mathrm{T}_{2}=272.9 \mathrm{~K}$

## TOTAL 4 points

$$
\begin{gathered}
\frac{v_{1} R T_{1}}{V}=\frac{v_{2} R T_{2}}{V} \\
v_{1} T_{1}=v_{2} T_{2} \\
T_{2}=v_{1} T_{1} / v_{2}=1.69 \cdot 373 / 2.31=272.9 \mathrm{~K}
\end{gathered}
$$

Determination of the temperature of the second thermostat - 2 points

## TASK № 8

In numbered test tubes, there are solutions of the following substances: zinc nitrate, barium nitrate, hydrochloric acid, sulfuric acid, and sodium carbonate. Without using other reagents, propose a method for identifying the substances. Write the equations for the necessary chemical reactions.

## Solution:

Tube numbers: №1 - $\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$, №2 - $\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}$, №3 - $\mathrm{HCl}, \mathrm{N} 04$ - $\mathrm{H}_{2} \mathrm{SO}_{4}$, №5 - $\mathrm{Na}_{2} \mathrm{CO}_{3}$.

Determination of tube numbers - 2 points

Precipitates formed: $\mathrm{BaSO}_{4}, \mathrm{BaCO}_{3}, \mathrm{ZnCO}_{3}$,

Precipitates $\mathrm{BaCO}_{3}, \mathrm{ZnCO}_{3}, \mathrm{Zn}(\mathrm{OH})_{2}$ dissolve in $\mathrm{HCl}$.

Determination of precipitates formed and their solubility - 2 points

Equations of reactions:

$\mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{HCl}=2 \mathrm{NaCl}+\mathrm{CO}_{2} \uparrow+\mathrm{H}_{2} \mathrm{O}$
$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}=\mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{CO}_{2} \uparrow+\mathrm{H}_{2} \mathrm{O}$

$\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}=\mathrm{BaSO}_{4} \downarrow+2 \mathrm{HNO}_{3}$

$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}=\mathrm{BaCO}_{3} \downarrow+2 \mathrm{NaNO}_{3}$

$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}=\mathrm{ZnCO}_{3} \downarrow+2 \mathrm{NaNO}_{3}$

Simultaneously with the formation of $\mathrm{ZnCO}_{3}$, a product of its hydrolysis - less soluble $\mathrm{Zn}(\mathrm{OH})_{2}$ - precipitates.

$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{H}_{2} \mathrm{O}=\mathrm{Zn}(\mathrm{OH})_{2} \downarrow+2 \mathrm{NaNO}_{3}+\mathrm{CO}_{2} \uparrow+\mathrm{H}_{2} \mathrm{O}$

Determination of reaction equations - 12 points

(2 points for each equation)

TOTAL 16 points

## Chemistry 10th grade  3rd VARIANT 2022-2023

## TASK № 1

A solution of sodium sulfite weighing 94.5 g with a mass fraction of the dissolved substance $15 \%$ was mixed with an excess of hydrochloric acid. Calculate the volume of the gas released.

## Solution:

11. Write the equation for the reaction between sodium sulfite and hydrochloric acid

$\mathrm{Na}_{2} \mathrm{SO}_{3}+\mathrm{HCl}=2 \mathrm{NaCl}+\mathrm{SO}_{2} \uparrow+\mathrm{H}_{2} \mathrm{O}$

Sign the amount of substance (in moles) under sodium sulfite and sulfur dioxide

$\mathrm{Na}_{2} \mathrm{SO}_{3}+\mathrm{HCl}=2 \mathrm{NaCl}+\mathrm{SO}_{2} \uparrow+\mathrm{H}_{2} \mathrm{O}$

1 mole

1 mole

Writing the reaction equation and balancing the coefficients - 2 points

12. Calculate the mass of sodium sulfite in the initial solution

$m\left(\mathrm{Na}_{2} \mathrm{SO}_{3}\right)=\frac{m(\mathrm{p}-\mathrm{pa}) \cdot \omega(\mathrm{v}-\mathrm{va})}{100 \%}=\frac{94.5 \cdot 15}{100}=14.175$ g

Determination of the mass of the substance - 2 points