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defon
3 years ago
12

WHICH OF THE GENOTYPES IN NUMBER 1 WOULD BE CONSIDERED PUREBRED

Chemistry
1 answer:
Rasek [7]3 years ago
3 0
Purebred<span> — Also called HOMOZYGOUS and consists of gene pairs with genes that are the SAME. Hybrid - Also called HETEROZYGOUS and consists of gene pairs that are D i'FFEREN'i". </span>Genotype<span> is the actual GENE makeup represented by LE'H'ERS.</span>
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The eight protons in oxygen atom's nucleus will exert more force than nitrogen's seven electrons. Also, oxygen has same number of shells as nitrogen. Thus electrons in oxygen atom will be closer to nucleus than in the case of nitrogen. When electrons are closer it means atomic radius is smaller.

Hope this helps! :)

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....... Is the energy accompanying the addition of electron to a gaseous atom.
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Which career best fits the group of words below?
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The answer is "power dispatcher."

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Why does the temperature of a matter doesn’t change during change of state upon heating?
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6 0
3 years ago
Freeze-drying is a process used to preserve food. If strawberries are to be freeze-dried, then they would be frozen to -80.00 °C
katrin2010 [14]

Answer:

1. 389 kJ; 2. 7.5 µg; 3. 6.25 days

Explanation:

1. Energy required

The water is converted directly from a solid to a gas (sublimation).

They don't give us the enthalpy of sublimation, but

\Delta_{\text{sub}}H = \Delta_{\text{fus}}H + \Delta_{\text{vap}}H = 6.01 + 40.68 = 46.69 \text{ kJ}\cdot\text{mol}^{-1}

The equation for the process is then

Mᵣ:                         18.02

         46.69 kJ + H₂O(s) ⟶ H₂O(g)

m/g:                       150

(a) Moles of water

\text{Moles} = \text{150 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{8.324 mol}

(b) Heat removed

46.69 kJ will remove 1 mol of ice.

\text{Heat removed} = \text{8.234 mol} \times \dfrac{\text{46.69 kJ}}{\text{1 mol}} = \textbf{389 kJ}\\\text{It takes $\large \boxed{\textbf{389 kJ}}$ to remove 150 g of ice}

2. Mass of water vapour in the freezer

For this calculation, we can use the Ideal Gas Law — pV = nRT

(a) Moles of water

Data:

p = 1.00 \times 10^{-3}\text{ torr } \times \dfrac{\text{1 atm}}{\text{760 torr}} = 1.316 \times 10^{-6}\text{ atm}

V = 5 L

T = (-80 + 273.15) K = 193.15 K

Calculation:

\begin{array}{rcl}pV & = & nRT\\1.316 \times 10^{-6}\text{ atm} \times \text{5 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{193.15 K }\\6.6 \times 10^{-6} & = & 15.85n\text{ mol}^{-1} \\n & = & \dfrac{6.6 \times 10^{-6}}{15.85\text{ mol}^{-1}}\\\\& = & 4.2 \times 10^{-7} \text{ mol}\\\end{array}

(b) Mass of water

\text{Mass} = 4.2 \times 10^{-7} \text{ mol} \times \dfrac{\text{18.02 g}}{\text{1 mol}} = 7.5 \times 10^{-6}\text{ g} = 7.5 \, \mu \text{g}\\\\\text{At any given time, there are $\large \boxed{\textbf{7.5 $\mu$g}}$ of water vapour in the freezer.}

3. Time for removal

You must remove 150 mL of water.

It takes 1 h to remove 1 mL of water.

\text{Time} = \text{150 mL} \times \dfrac{\text{1 h}}{\text{1 mL}} = \text{150 h} = \text{6.25 days}

5 0
3 years ago
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