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3 years ago

For species that reproduce asexually, all their offspring are

Chemistry

1 answer:

inna [77]3 years ago

8 0

Answer:

A. genetically identical

Explanation:

Asexual reproduction produces offspring that are genetically identical to the parent because the offspring are all clones of the original parent. A single individual can produce offspring asexually and large numbers of offspring can be produced quickly

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How many TOTAL atoms are contained in 10 molecules of TNT, C7H5N3O6?

nlexa [21]

Answer:7 atom of carbon+5atom of hydrogen +3atom of nitrogen +6 atom of oxygen =21

5 0

3 years ago

3. 4 g of a nonelectrolyte dissolved in 78. 3 g of water produces a solution that freezes at −4. 5∘c. what is the molar mass of

enyata [817]

3. 4 g of a nonelectrolyte dissolved in 78. 3 g of water produces a solution. The molar mass of the solute will be 17.94.

<h3>What is molar mass?</h3>

Molar mass of a substance is its mass in grams in per mole of a solution.

Freezing point: Freezing point of a substance is a temperature at which a liquid starts to solidify.

Depression in the freezing point can be calculated

[Depression in freezing point of pure solvent—Freezing point of solution] =[(0) - (-4.5)] °C =4.5 °C

molar mass = Number of moles of solute m / Mass of solvent in Kg

3.4g / M x 1/ 0.0783 kg = 43.42

Substitute AT by 4.5°C , Kr by 1.86 °C/m, and m by 43.42 m in equation (1) as follows:

1.86 x 43.42 / 4.5 = 17.94

Therefore, molar mass of solute to be 17.94.

To learn more about molar mass, refer to the link:

brainly.com/question/22997914

#SPJ4

4 0

2 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago

What type of weather is associated with a continental polar air mass?What is a front?

tester [92]

Answer:

I would guess that a polar air mask would be freezing, high winds type weather. I think that's right.

8 0

4 years ago

Four (4.0) liters of helium gas are stored at 125 atm pressure. If the temperature and number of particles do not change, what w

seropon [69]

Answer:

10 L

Explanation:

The only variables are pressure and volume, so we can use Boyle's Law:

p1V1 = p2V2

Data:

p1 = 125 atm; V1 = 4.0 L

p2 = 50 atm; V2 = ?

Calculation:

125 × 4.0 = 50V2

500 = 50 V2

V2 = 500/50 = 10 L

The new volume will be 10 L.

5 0

3 years ago