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2 years ago

_______is the division of the cytoplasm at the end of mitosis.

Chemistry

1 answer:

tankabanditka [31]2 years ago

6 0

Answer: D. CYTOKINESIS

Explanation:

Mitosis ends with telophase, or the stage at which the chromosomes reach the poles. ... Telophase is followed by CYTOKINESIS, or the division of the cytoplasm into two daughter cells.

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1826.5g of methanol (CH3OH), molar mass = 32.0 g/mol is added to 735 g of water, what is the molality of the methane 0.0348 m 1.

Snowcat [4.5K]

Answer:

Molality = 1.13 m

Explanation:

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of $CH_3OH$ = 26.5 g

Molar mass of $CH_3OH$ = 32.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{26.5\ g}{32.04\ g/mol}$

$Moles\ of\ CH_3OH= 0.8271\ moles$

Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )

So, molality is:

$m=\frac {0.8271\ moles}{0.735\ kg}$

Molality = 1.13 m

4 0

3 years ago

What occurs when potassium reacts with chlorine to form potassium chloride?

schepotkina [342]

The potassium will donate one of its valence electrons

6 0

2 years ago

Read 2 more answers

How much heat is required to vaporize 31.5 gg of acetone (C3H6O)(C3H6O) at 25 ∘C∘C? The heat of vaporization for acetone at this

KiRa [710]

Answer:

≅ 16.81 kJ

Explanation:

Given that;

mass of acetone = 31.5 g

molar mass of acetone = 58.08 g/mol

heat of vaporization for acetone = 31.0 kJ/molkJ/mol.

Number of moles = $\frac{mass}{molar mass}$

Number of moles of acetone = $\frac{31.5}{58.08}$

Number of moles of acetone = 0.5424 mole

The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;

Hence;

The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol

The heat required to vaporize 31.5 g of acetone = 16.8144 kJ

≅ 16.81 kJ

4 0

3 years ago

A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,

choli [55]

Answer:

3.676 L.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have different values of V and T:

(V₁T₂) = (V₂T₁)

Knowing that:

V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,

V₂ = ??? L, T₂ = 40°C + 273 = 313 K,

Applying in the above equation

(V₁T₂) = (V₂T₁)

∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.

7 0

3 years ago

A solution is made by dissolving 3.35 g of fructose in 35.0 ml of water. What is the molality of fructose in the solution? Assum

ludmilkaskok [199]

Answer:

m = 0.531 molal

Explanation:

∴ m fructose = 3.35 g

∴ V water = 35.0 mL

∴ ρ H2O = 1 g/mL

molality = moles solute / Kg solvent

∴ Mw fructose = 180.16 g/mol

⇒ moles fructose = 3.35 g * ( mol / 180.16 g) = 0.0186 mol fructose

⇒ m H2O = 35.0 mL * ( 1 g/mL ) * ( Kg/1000g) = 0.035 Kg H2O

⇒ molality (m) = 0.0186 mol fructose / 0.035 Kg H2O

⇒ m = 0.531 molal

6 0

3 years ago