Which element's atoms have a larger covalent

In run 1, you mix 7.9 mL of the 43 g/L MO solution (MO molar mass is 327.33 g/mol), 3.13 mL of the 0.040 M SnCl2 in 2.0 M HCl so

blsea [12.9K]

Answer:

Concentration of H3O⁺ [H3O⁺] = 0.864 M

Explanation:

Given that:

The mass concentration of MO = 43 g/L

The volume of MO = 7.9 mL = 7.9 × 10⁻³ L

Recall that

The mass number of MO = Mass concentration of MO × Volume of MO

The mass number of MO = (43 g/L) * (7.9 × 10⁻³ L)

The mass number of MO = 0.3397 g

number of moles of MO = (mass number of MO) / (molar mass of MO)

number of moles of MO = (0.3397 g) / (327.33 g/mol)

moles of MO = 0.00104 mol

The total volume = 7.9 mL + 3.13 mL + 5.49 mL + 3.43 mL

The total volume = 19.95 mL = 19.95 × 10⁻³ L

Concentration of MO [MO} =(number of moles of MO) / (total volume)

[MO] = 0.00104 mol / 19.95 × 10⁻³ L

[MO] = 5.2130 × 10⁻⁸ M

the number of moles of H3O⁺ = molarity of HCl in the solution × the volume of HCl in solution

the number of moles of H3O⁺ = [(2.0 M) * (3.13 mL)] + [(2.0 M) * (5.49 mL)]

the number of moles of H3O⁺ = 17.24 mmol

Concentration of H3O⁺ [H3O⁺] = (the number of moles of H3O⁺) / (total volume)

Concentration of H3O⁺ [H3O⁺] = (17.24 mmol) / (19.95 mL)

Concentration of H3O⁺ [H3O⁺] = 0.864 M

5 0

3 years ago

Which of these structures is found in all types of plant and animal cells

liq [111]

Answer:

Hi, There!

which of these structures is found in all types of plant and animal cells

Cell membrane is found in both animal and plant cells.

Nucleus is found in both animal and plant cells.

Cytoplasm is found in both animal and plant cells.

Mitochondria is found in both animal and plant cells.

Endoplasmic reticulum is found in both animal and plant cells.

Golgi apparatus is found in both animal and plant cells.

Lysosomes is found in both animal and plant cells.

$~Hope ~ this ~ Helps~$

5 0

3 years ago

9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1

WITCHER [35]

Answer:

$Keq=11.5$

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

$CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)$

We can write the law of mass action as:

$Keq=\frac{[CH_3OH]}{[CO][H_2]^2}$

That in terms of the change $x$ due to the reaction extent we can write:

$Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}$

Nevertheless, for the carbon monoxide, we can directly compute $x$ as shown below:

$[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\$

$[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\$

$[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\$

$x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M$

Finally, we can compute the equilibrium constant:

$Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5$

Best regards.

3 0

4 years ago

Suppose 20.23 g of glucose are dissolved in 95.75 g of water at 27.0 OC. Glucose is nonvolatile (has no vapor pressure) and has

leonid [27]

Answer:

Explanation:

From the information given :

we can understand the solute is glucose and the solvent is water,

So, the weight of glucose = 20.23 g

the molecular weight of glucose = 180.2 g/mol

weight of water = 95. 75 g

the molecular weight of water = 18.02 g/mol

pure vapor pressure of water $P_A = 26.7 \ mmHg$ at 27°C

moles of glucose = weight of glucose/ molecular weight of glucose

= 20.23/180.2

= 0.11 mole

moles of water = weight of water / molecular weight of water

= 95.75/18.02

= 5.31 mole

mole fraction of glucose $X_{glucose} =$ (moles of glucose)/(moles of glucose+ moles of water)

$X_{glucose} =$ 0.11/(0.11 + 5.31)

$X_{glucose} =$ 0.0203

mole fraction of glucose $X_{water} =$ (moles of water)/(moles of water+ moles of glucose)

$X_{water} =$ 5.31/ (5.31 + 0.11)

$X_{water} =$ 0.9797

Using Raoult's Law:

$P_S = P^0_A \times X_A \ \ \ OR \ \ \ P_A = P^0_A \times X_A$

where:

$P_S$ = vapor pressure of the solution

$P_A$ = total vapor pressure of the solution

$P^0_A$ = vapor pressure of the solvent in the pure state

$X_A$ = mole fraction of solvent i.e. water

$P_A =$ 95.75 × 0.9797

$P_A =$ 93.81 mmHg

the total vapor pressure of the solution = 93.81 mmHg

4 0

3 years ago

How would an increase in temperature affect the rate of radioactive decay?

bulgar [2K]

Answer:

Temperature does not affect the radioactive decay

Explanation:

because Temperature has no factor in radioactivity

Good Luck

8 0

3 years ago