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Andrews [41]
3 years ago
14

What is a change in the arrangement of bases in a gene called?????

Chemistry
2 answers:
jek_recluse [69]3 years ago
5 0
Asexual reproduction
Jobisdone [24]3 years ago
4 0
Asexual reproduction.<span>
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You might be interested in
4 g of ag2so4 will dissolve in 1l of water. calculate the solubility product (ksp) for silver (i) sulfate.
Bas_tet [7]
Reaction of dissociation: Ag₂SO₄ → 2Ag⁺ + SO₄²⁻.
m(Ag₂SO₄) = 4 g.
V(Ag₂SO₄) = 1 l.
n(Ag₂SO₄) = m(Ag₂SO₄) ÷ M(Ag₂SO₄).
n(Ag₂SO₄) = 4 g ÷ 311,8 g/mol.
n(Ag₂SO₄) = 0,0128 mol.
n(Ag⁺) = 2 · 0,0128 mol = 0,0256 mol.
n(Ag₂SO₄) = n(SO₄²⁻) = 0,0128 mol.
c(Ag⁺) = n ÷ V = 0,0256 mol ÷ 1 l = 0,0256 mol/l.
Ksp = c(Ag⁺)² · c(SO₄²⁻).
Ksp = (0,0256 mol/l)² · 0,0128 mol/l.
Ksp = 8,3·10⁻⁶.

7 0
3 years ago
A 32.4 L gas sample at STP is compressed to a volume of 28.4 L, and the temperature is increased to 352 K. What is the new press
Sindrei [870]

Answer:

1.47 atm

Explanation:

Step 1: Given data

  • Initial volume (V₁): 32.4 L
  • Initial pressure (P₁): 1 atm (standard pressure)
  • Initial temperature (T₁): 273 K (standard temperature)
  • Final volume (V₂): 28.4 L
  • Final pressure (P₂): ?
  • Final temperature (T₂): 352 K

Step 2: Calculate the final pressure of the gas

We can calculate the final pressure of the gas using the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

P₂ = P₁ × V₁ × T₂ / T₁ × V₂

P₂ = 1 atm × 32.4 L × 352 K / 273 K × 28.4 L = 1.47 atm

7 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
Which model shows a molecule of a substance that is made up of three elements ​
vovangra [49]

Answer:

c or d

Explanation:

i tried a and it was wrong. its not B. and i cant see the rest. its the one that looks like a catipillar tho

4 0
2 years ago
What volume of sample, in mL, would be needed to make 2 mL of a 5-fold dilution? (3 significant figures needed)
natulia [17]

Answer:

0.400 mL

Explanation:

Hello, the dilution factor (in folds) is given by:

folds=\frac{total_{volume}}{sample_{volume}}

Thus, the sample volume with three significant figures is given by:

sample_{volume}=\frac{2mL}{5"folds"}=0.400 mL

Best regards.

5 0
3 years ago
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