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Ainat [17]
3 years ago
8

1. When a broken stem grows roots, ( blank ) reproduction takes place.

Chemistry
1 answer:
Sliva [168]3 years ago
3 0
I believe the answers would be...
1. Vegetative
2. True
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Type the correct answer in the box. Express your answer to three significant figures.
satela [25.4K]

Answer:

The partial pressure of argon in the jar is 0.944 kilopascal.

Explanation:

Step 1: Data given

Volume of the jar of air = 25.0 L

Number of moles argon = 0.0104 moles

Temperature = 273 K

Step 2: Calculate the pressure of argon with the ideal gas law

p*V = nRT

p = (nRT)/V

⇒ with n = the number of moles of argon = 0.0104 moles

⇒ with R = the gas constant = 0.0821 L*atm/mol*K

⇒ with T = the temperature = 273 K

⇒ with V = the volume of the jar = 25.0 L

p = (0.0104 * 0.0821 * 273)/25.0

p = 0.00932 atm

1 atm =101.3 kPa

0.00932 atm = 101.3 * 0.00932 = 0.944 kPa

The partial pressure of argon in the jar is 0.944 kilopascal.

5 0
4 years ago
A student prepared a stock solution by dissolving 10.0 g of KOH in enough water to make 150. mL of solution. She then took 15.0
yanalaym [24]

Answer: The concentration of KOH for the final solution is 0.275 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

 V_s = volume of solution in ml = 150 ml

moles of solute =\frac{\text {given mass}}{\text {molar mass}}=\frac{10.0g}{56g/mol}=0.178moles

Now put all the given values in the formula of molality, we get

Molality=\frac{0.178\times 1000}{150}=1.19M

According to the dilution law,

C_1V_1=C_2V_2

where,

C_1 = molarity of stock solution = 1.19 M

V_1 = volume of stock solution = 15.0 ml

C_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 65.0 ml

Putting in the values we get:

1.19\times 15.0=M_2\times 65.0

M_1=0.275M

Therefore, the concentration of KOH for the final solution is 0.275 M

5 0
3 years ago
A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.
dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution =  9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

3 0
3 years ago
The combustion of 0.25 mol of an unknown organic compound results in the release of 320 kJ of energy. Which of the compounds in
vichka [17]

Answer: C. ethanol

The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.

<u>The enthalpy of combustion of the unknown compound is</u>

ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol

<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,

e% =  ( | ΔHx - ΔH | / ΔHx ) x 100%

where ΔHx is the enthalpy of combustion of the probable compound.

The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol

Compound Enthalpy of combustion (kJ/mol)   Deviation

Methane                        - 890.7                                 43.8%

Ehylene                         -1411.2                                   9.3%

Ethanol                        -1368.6                                    6.5%

According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.

5 0
4 years ago
Read 2 more answers
Why must the spot applied to a TLC plate be above the level of the developing solvent? What problem will ensue if the level of t
algol13

Answer:

See explanation

Explanation:

If the spot in TLC is below the solvent front, it will be observed that the spot, instead of being separated by the solvent as expected, will just dissolve away in the solvent and zero actual separation of the mixture is achieved.

If the solute is dissolved away instead of being separated by the solvent, then the experiment fails because no actual separation of the mixture is achieved.

Hence, in TLC, the spot must be applied above the solvent front so that the capillary movement of the solvent through the plate can lead to the eventual separation of the components of the mixture since the various components of the mixture will travel at different speeds through the plate.

Also, if the solvent is above the spot, the solvent may evaporate selectively from the points above the spot while separation is ongoing.

6 0
3 years ago
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