This group has characteristics including: brittleness, insulators/poor conductors, and are often found as gases.

7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?

Illusion [34]

Answer:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

3 0

2 years ago

William measures a test tube and finds that the mass of the test tube is 5g. He places the lone reactant in the test tube and fi

iogann1982 [59]

86 percent is the percent yield for this experiment if he expected to produce 5g of product.

Explanation:

Given that:

mass of test tube = 5 grams

mass of test tube + reactant is 12.5 grams

mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)

mass of reactant = 12.5 -5

= 7.5 grams

when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.

so mass of product formed = 9.3 - 5

= 4. 3 grams of product is formed (actual yield)

However, he expected the product to be 5 grams (theoretical yield)

Percent yield = $\frac{actual yield}{theoretical yield}$ x 100

putting the values in the formula:

percent yield = $\frac{4.3}{5}$ x 100

= 86 %

86 percent is the percent yield.

5 0

3 years ago

According to an informal 1992 survey, the drinking water in about one-third of the homes in Chicago had lead levels of about 10

wel

Answer:

10.2 mg

Explanation:

Step 1: Calculate the total amount of water she drank

1 year has 365 days and she lived in Chicago for 2 years = 2 × 365 days = 730 days.

If she drank 1.4 L of water per day, the total amount of water she drank is:

730 day × 1.4 L/day = 1022 L

Step 2: Calculate the amount of Pb in 1022 L of water

The concentration of Pb is 10 ppb (10 μg/L).

1022 L × 10 μg/L = 10220 μg

Step 3: Convert 10220 μg to milligrams

We will use the conversion factor 1 mg = 1000 μg.

10220 μg × 1 mg/1000 μg = 10.2 mg

8 0

2 years ago

How are the environments of a desert and a tundra different?

klasskru [66]

B. A Tundra is less humid than a desert.

Reasoning: Tundra’s experience low temperatures while deserts experience hot/humid temperatures.

6 0

2 years ago

Three moles of an ideal gas undergo a reversible isothermal compression at temperature 18.0 ?c. during this compression, an amou

Lena [83]

We calculate the entropy of an ideal gas follows:

For an isothermal compression, change in internal energy is equal to zero.
Thus, the heat added to the gas is equal to the work done on the gas which is given as 1750 J.

Entropy would be 1750/301 = 5.81 J/K

3 0

3 years ago