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balandron [24]
3 years ago
12

To compare how different types of soil absorbed water, Christian set up an experiment. He poked equal numbers and sizes of holes

into the bottoms of three plastic cups. He filled one cup with sand, one with potting soil, and left one empty. He placed each cup in a stand over a beaker. Christian poured 400 mL (milliliters) of water into each of the cups. He then measured the amount of water collected in the beakers beneath each cup at 1 minute intervals for 3 minutes. What is the control in this experiment? Group of answer choices
Chemistry
1 answer:
lilavasa [31]3 years ago
3 0

Answer:

The empty cup

Explanation:

Got it right on the test

You might be interested in
How does Earth's rotation affect ocean currents?
Tom [10]

Answer:

Our planet's rotation produces a force on all bodies moving relative to theEarth. ... The force, called the "Coriolis effect," causes the direction of winds and ocean currents to be deflected.

Explanation:

4 0
2 years ago
A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th
tatyana61 [14]

Answer : The partial pressure of the CO_2 in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % CO_2 and 25 % N_2 in terms of volume.

First we have to calculate the moles of CO_2 and N_2.

\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole

\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole

Now we have to calculate the mole fraction of CO_2.

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}

\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75

Now we have to calculate the partial pressure of the CO_2 gas.

\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}

\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the CO_2 in the tank in psia is, 32.6 psia.

3 0
3 years ago
Science 10 workbook
Kruka [31]

Answer:

for more details are in the pic

4 0
3 years ago
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
How many particles of Na are there in 1.43g of a molecular compound with a Molar mass of 23g?
Olin [163]

Answer:

3.74 x 10²² particles

Explanation:

Given parameters:

Mass of compound  = 1.43g

Molar mass of compound  = 23g

Unknown:

Number of particles of sodium = ?

Solution:

To find the number of particles of Na in the compound, we need to obtain the mass of sodium from the total mass given;

          Mass of sodium  = \frac{molar mass of Na}{molar mass of compound} x mass of sample

                                      = \frac{23}{23}  x 1.43g

                                       = 1.43g

Now find the number of moles of this amount of Na in the sample;

          Number of moles  = \frac{mass}{molar mass} = \frac{1.43}{23}  = 0.062mole

Now;

                    1 mole of substance  = 6.02 x 10²³ particles

                       0.062 mole of substance  =  0.062 x 6.02 x 10²³ particles

                                                                     = 3.74 x 10²² particles

7 0
3 years ago
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