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4 years ago

What prevents the expression of a recessive allele

2 answers:

7 0

Usually the presence of a dominant version of allele masks the expression of a recessive allele. A recessive allele is only phenotype is only expressed when 2 recessive alleles combine.

astra-53 [7]4 years ago

3 0

Dominant alleles prevent the expression of a recessive allele, unless it is a genotype of two recessive alleles.

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Describe how you would prepare 100 mL of a 0.050 M sodium citrate tribasic buffer to a pH of 6. This method is accomplished by p

Vinvika [58]

Answer:

To prepare 100 mL of a 0,050M sodium citrate tribasic buffer to a pH of 6 you need to add 7,16 mL of 0,5M HCl, 1,4705 g of sodium citrate tribasic dihydrate and complete 100 mL with water.

Explanation:

The acid equilibrium of sodium citrate tribasic buffer is:

citrate dibasic⁻² ⇄ citrate tribasic⁻³ + H⁺ pka = 6,4

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ $\frac{[A^{-}]}{[HA]}$

6,0 = 6,4 + log₁₀ $\frac{[CitrateTribasic]}{[CitrateDibasic]}$

0,3981 = $\frac{[CitrateTribasic]}{[CitrateDibasic]}$ (1)

You need to add 0,1L× 0,050M = 0,0050moles of sodium citrate:

0,0050 moles = Citrate tribasic + Citrate dibasic (2)

Replacing (2) in (1)

Citrate dibasic: 3,58x10⁻³ moles

Thus,

Citrate tribasic: 1,42x10⁻³ moles

Citrate tribasics reacts with HCl thus:

Citrate tribasic⁻³ + HCl → Citrate dibasic⁻² + Cl⁻

Thus, you need to add 5x10⁻³moles of sodium citrate tribasic and 3,58x10⁻³ moles of HCl:

5x10⁻³moles of sodium citrate tribasic× $\frac{294,1 g}{1mole}$ = 1,4705 g of sodium citrate tribasic dihydrate -commercial reactant-

3,58x10⁻³ moles of HCl÷ 0,5 M = 7,16x10⁻³ L ≡ 7,16 mL of 0,5M HCl

Thus, to prepare 100 mL of a 0,050M sodium citrate tribasic buffer to a pH of 6 you need to add 7,16 mL of 0,5M HCl, 1,4705 g of sodium citrate tribasic dihydrate and complete 100 mL with water.

6 0

4 years ago

Calculate the energy required to heat 1.30kg of water from 22.4°C to 34.2°C . Assume the specific heat capacity of water under t

Serhud [2]

Answer:

The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J

Explanation:

Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.

The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:

Q = c * m * ΔT

Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).

In this case:

$c=4.18 \frac{J}{g*K}$

m= 1.30 kg= 1,300 g (1 kg=1,000 g)
ΔT= 34.2 °C - 22.4 °C= 11.8 °C= 11.8 °K Being a temperature difference, it is independent if they are degrees Celsius or degrees Kelvin. That is, the temperature difference is the same in degrees Celsius or degrees Kelvin.

Replacing:

$Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K$

Q= 64,121.2 J

The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J

4 0

3 years ago

Steps of scientific inquiry?

vlabodo [156]

1. Figure out a problem
2. Research
3. Hypothesis
4. Experiment
5. Analyze results
6. Conclusion
7. Communicate your results

Hope This Helps. :D

8 0

3 years ago

Read 2 more answers

A volume of 120 mL of H2O is initially at room temperature (22.00 ∘C ). A chilled steel rod at 2.00 ∘C ∘C is placed in the water

aliina [53]

Answer:

$\large \boxed{\text{28.5 g}}$

Explanation:

There are two heat flows in this process and, since energy (heat) can neither be destroyed nor created, the energy change for the system must equal zero.

Data:

For Fe, m₁ = ?; C₁ = 0.452 J°C⁻¹g⁻¹; Ti = 2.00 °C; T_f = 21.50 °C

For H₂O, m₂ = 120 g; C₂ = 4.18 J°C⁻¹g⁻¹; Ti = 22.00 °C; T_f = 21.50 °C

Calculations:

1. Temperature changes

ΔT₁ = T_f - Ti = 21.50 °C - 2.00 °C = 19.50 °C

ΔT₂ = T_f - Ti = 21.50 °C - 22.00 °C = -0.50 °C

2. Mass of steel rod

$\begin{array}{ccl}\text{Heat gained by steel rod + heat lost by water} & = & 0\\m_{1}C_{1} \Delta T_{1} + m_{2}C_{2} \Delta T_{2}& = & 0\\m_{1} \times 0.452 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\times 19.50 \, ^{\circ}\text{C} + \text{120 g} \times 4.18\text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times (-0.50)\, ^{\circ}\text{C}& = & 0\\8.814m_{1}\text{ g}^{-1} - 250.8 & = &0\\\end{array}\\$

$\begin{array}{ccl}8.814m_{1}\text{ g}^{-1} & = &250.8\\m_{1} & = & \dfrac{250.8}{\text{8.814 g}^{-1}}\\\\& = & \textbf{28.5 g}\\\end{array}\\\text{The mass of the steel rod is $\large \boxed{\textbf{28.5 g}}$}$

6 0

3 years ago

Que estudia la química

Sunny_sXe [5.5K]

Answer:

no hablo espanol ok? sorry

6 0

3 years ago