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Lyrx [107]
2 years ago
15

the following punnett square shows a cross between two hybrid flowers. if the flowers show incomplete dominance, what fraction a

re pure red? I NEED HELP PLZ!

Chemistry
2 answers:
Len [333]2 years ago
6 0

Answer:

1/4

Explanation:

Incomplete dominance is when the dominant ("R") gene is not completely dominant and cannot fully mask or cover the recessive ("r") gene. This means that it will result in both genes expressing and a mix of traits appearing. In the case of a flower with "R" genes for red and "r" genes for white, the following possibly combinations exist:

RR = all red, no white "r" gene

Rr = pink, as both the "R" and "r" genes are expressed

rr = white, no red "R" gene

The Punnett Square shows that only one of the four phenotypes is RR and red.

djverab [1.8K]2 years ago
4 0
I think it’s 1/4!!!!
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It takes 2,500,00 Liters of Helium to fill the Goodyear Blimp. How many moles is this?
Valentin [98]

Answer:

102.26 moles of helium were required to Fill the Goodyear Blimp

Explanation:

To solve this question we need to use combined gas law:

PV = nRT

<em>Where P is pressure, V is volume of gas (2500L), n are moles of gas (Our incognite), R is gas constant (0.082atmL/molK) and T is absolute temperature</em>

<em />

Assuming atmospheric condition we can write P = 1atm and T = 25°C = 298.15K

Replacing:

PV/RT = n

1atm*2500L / 0.082atmL/molK*298.15K = n

<h3>102.26 moles of helium were required to Fill the Goodyear Blimp</h3>

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3 0
2 years ago
PLEASE HELP this is physical science.
Andru [333]

Bone age : 22,920 years

<h3>Further explanation</h3>

Given

Nt = 2.5 g C-14

No = 40 g

half-life = 5730 years

Required

time of decay

Solution

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

Input the value :

\tt 2.5=40.\dfrac{1}{2}^{t/5730}\\\\\dfrac{2.5}{40}=\dfrac{1}{2}^{t/5730}\\\\(\dfrac{1}{2})^4=\dfrac{1}{2}^{t/5730}\\\\4=t/5730\rightarrow t=22920~years

6 0
2 years ago
Priscilla was building a circuit that used copper wires to connect a battery to a light bulb. As she connected the final wire fr
algol [13]

Answer:

The voltage or potential difference

Explanation:

What makes current flow in a circuit is the voltage or the potential difference.

This force is supplied by the battery or the mains electrical circuit.

  • Every circuit requires the voltage to drive current through
  • When a circuit is complete, the battery is able to overcome any resistance by the generating enough voltage which is the force to drive the current through.
7 0
2 years ago
How is density related to the color of igneous rocks?​
r-ruslan [8.4K]

Answer:

The density of igneous rocks is related to its color. Darker colored rocks have a higher density because of its greater mineral and iron content. Its characteristics is opposite compared to lighter colored rocks that have less density because of lower mineral and iron content

4 0
2 years ago
In IR spectroscopy, we normally talk about "frequencies" when in reality we are referring to wavenumbers. What is the mathematic
Svetach [21]

Answer:

Here's what I get.

Explanation:

(b) Wavenumber and wavelength

The wavenumber is the distance over which a cycle repeats, that is, it is the number of waves in a unit distance.

\bar \nu = \dfrac{1}{\lambda}

Thus, if λ = 3 µm,

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(a) Wavenumber and frequency

Since

λ = c/f and 1/λ = f/c

the relation between wavenumber and frequency is

\bar \nu = \mathbf{\dfrac{f}{c}}

Thus, if f = 90 THz

\bar \nu = \dfrac{90 \times 10^{12} \text{ s}^{-1}}{3 \times 10^{8} \text{ m$\cdot$ s}^{-1}}= 3 \times 10^{5} \text{ m}^{-1} = \textbf{3000 cm}^{-1}

(c) Units

(i) Frequency

The units are s⁻¹ or Hz.

(ii) Wavelength

The SI base unit is metres, but infrared wavelengths are usually measured in micrometres (roughly 2.5 µm to 20 µm).

(iii) Wavenumber

The SI base unit is m⁻¹, but infrared wavenumbers are usually measured in cm⁻¹ (roughly 4000 cm⁻¹ to 500 cm⁻¹).

8 0
3 years ago
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