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3 years ago

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Chemistry

2 answers:

Marizza181 [45]3 years ago

8 0

Answer:

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Explanation:

bsbsbsnsnsnsjs

Irina-Kira [14]3 years ago

3 0

Answer:

sorry I don't know this language

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A compound contains sulfur, oxygen, and chlorine. Analysis shows that it contains by mass 26.95% sulfur and 59.61% chlorine. Wha

sladkih [1.3K]

Answer:

SCl₂

Explanation:

In order to know the empirical formula, we have to follow a series of steps.

Step 1: Divide each percentage by the atomic mass

S: 26.95/32.07 = 0.8403

Cl: 59.61/35.45 = 1.682

Step 2: Divide all the numbers by the smallest one.

S: 0.8403/0.8403 = 1

Cl: 1.682/0.8403 ≈ 2

The empirical formula of the compound is SCl₂.

7 0

3 years ago

Read 2 more answers

Explain how the presence of several oxygen atoms ina compound containing an-OH group can make the compound acidic.

Lina20 [59]

In each case, the same bond gets broken - the bond between the hydrogen and oxygen in an -OH group. Writing the rest of the molecule as "X"

The factors to consider

Two of the factors which influence the ionisation of an acid are:

the strength of the bond being broken,the stability of the ions being formed.

In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to be similar.

7 0

3 years ago

An excess of oxygen reacts with 451.4 g of lead, forming 374.7 g of lead(II) oxide. Calculate the percent yield of the reaction.

Stels [109]

Answer: The percent yield of the reaction is 77.0 %

Explanation:

$2Pb+O_2\rightarrow 2PbO$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of lead}=\frac{451.4g}{207.2g/mol}=2.18moles$

$\text{Number of moles of lead oxide}=\frac{374.7g}{223.2g/mol}=1.68moles$

According to stoichiometry:

2 moles of $Pb$ produces = 2 moles of $PbO_2$

2.18 moles of $Pb$ is produced by= $\frac{2}{2}\times 2.18=2.18moles$ of $PbO_2$

Mass of $PbO_2$ = $moles\times {\text {Molar mass}}=2.18\times 223.2g/mol=486.6$

percent yield = $\frac{374.7g}{486.6g}\times 100=77.0\%$

3 0

3 years ago

Determine the freezing point of a 0.51 molal

Elan Coil [88]

Answer:

-65.897°C.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation: ΔTf = Kf.m,

where, ΔTf is the depression in freezing point of chloroform solution.

Kf is the molal depression constant of chloroform (Kf = 4.70°C.kg/mol).

m is the molality of the solution (m = 0.51 m).

∴ ΔTf = Kf.m = (4.70°C.kg/mol)(0.51 m) = 2.397°C.

∴ The freezing point of the solution = (freezing point of chloroform) - ΔTf = (-63.5°C) - (2.397°C) = -65.897°C.

7 0

3 years ago

A hot air balloon is filled with 1.31 × 10 6 L of an ideal gas on a cool morning ( 11 ∘ C ) . The air is heated to 121 ∘ C . Wha

victus00 [196]

Answer:

$1.82\times 10^6 L$ is the volume of the air in the balloon after it is heated.

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$ (at constant pressure)

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1= 1.31\times 10^6 L\\T_1=11^oC=(11+273.15)K=284.15K\\V_2=?\\T_2=121^oC=(121+273.15)K=394.15 K$

Putting values in above equation, we get:

$\frac{1.31\times 10^6 L}{284.15 K}=\frac{V_2}{394.14 K}\\\\V_2=\frac{V_1\times T_2}{T_1}$

$V_2=1.82\times 10^6 L$

$1.82\times 10^6 L$ is the volume of the air in the balloon after it is heated.

4 0

3 years ago

Срочно!!!!!!!помогите пожалуйста ​

Срочно!!!!!!!помогите пожалуйста