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3 years ago

EXPLAIN AND I GIVE BRAINLEST PLSSSSSSSSSSSSSSSSSSS GOOD EXPLAIN PLSSSS

Chemistry

2 answers:

prisoha [69]3 years ago

8 0

50% i don’t really know how to explain but i remover doing this in 7th grade

Alecsey [184]3 years ago

3 0

Answer:

Explanation:

In the Punnett Square, the first square would be RR, then on the right it would be Rr, the bottom left square would be Rr and the bottom right would be rr.

Therefore 25% is guaranteed to be straight rooted (RR) and 25%(rr) will be bent. The remaining 50% (Rr) could be either.

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One crystalline form of silica (SiO2) has a cubic unit cell, and from X-ray diffraction data it is known that the cell edge leng

tigry1 [53]

Answer:

There are 8Si atoms and 16 O atoms per unit cell

Explanation:

From the question we are told that:

Edge length $l=0.700nm=>0.7*10^9nm$

Density $\rho=2.32g/cm^3$

Generally the equation for Volume is mathematically given by

$V=l^3$

$V=(0.7*10^9)m^3$

$V=3,43*10^-{22}cm$

Where

Molar mass of (SiO2) for one formula unit

$M=28+32$

$M=60g/mol$

Therefore

Density of Si per unit length is

$\rho_{si}=\frac{9.96*10^{23}}{3.43*10^22}$

$\rho=0.29$

Molar mass of (SiO2) for one formula unit

$M=28+32$

$M=60g/mol$

Therefore

There are 8Si atoms and 16 O atoms per unit cell

6 0

4 years ago

Consider a 125 mL buffer solution at 25°C that contains 0.500 mol of hypochlorous acid (HOCl) and 0.500 mol of sodium hypochlori

Helga [31]

Answer:

pH = 6.82

Explanation:

To solve this problem we can use the Henderson-Hasselbach equation:

pH = pKa + log $\frac{[NaOCl]}{[HOCl]}$

We're given all the required data to calculate the original pH of the buffer before 0.341 mol of HCl are added:

pKa = -log(Ka) = -log(2.9x10⁻⁸) = 7.54

[HOCl] = [NaOCl] = 0.500 mol / 0.125 L = 4 M

pH = 7.54 + log $\frac{4}{4}$

pH = 7.54

By adding HCl, we simultaneously increase the number of HOCl and decrease NaOCl:

pH = 7.54 + log $\frac{[NaOCl-HCl]}{[HOCl+HCl]}$

pH = 7.54 + log $\frac{(0.500mol-0.341mol)/0.125L}{(0.500mol+0.341mol)/0.125L}$

pH = 6.82

6 0

3 years ago

Which of the following non-hydrogen atom transitions does the photon have at its long wavelength?

suter [353]

Answer:

The move from Level n=3 to Level n=2 has the long wavelength.

Explanation:

First, due to the selection rules, only transitions between adjacent levels are allowed, thus, only a transition between Level n=3 to Level n=2 or Level n=5 to Level n=4 are allowed. The two first options are wrong.

Second, analyzing the transition between Level n=3 to Level n=2 and the transition between Level n=5 to Level n=4 it is necessary to think in terms of the equation of the difference of energy for these type of transitions:

Δ $E = \frac{h^{2}}{8.m.L}(n_{LUMO}^{2} -n_{HOMO}^{2} )$ (1)

The difference in energy (ΔE) is directly proportional to the quadratic difference between the 'n' levels of transition. Therefore, If the transition occurs between smaller 'n' levels the difference of energy will be smaller too.

Also, the energy (ΔE) is inversely proportional to the wavelength (λ) so a smaller energy means a larger wavelength.

ΔE = c / λ (2)

Hence, the move from Level n=3 to Level n=2 has a long wavelength.

In order to calculate this wavelength is necessary to replace the data on equation (1) and (2).

6 0

4 years ago

the solubility product Ag3PO4 is: Ksp = 2.8 x 10^-18. What is the solubility of Ag3PO4 in water, in moles per liter?

guapka [62]

Answer : The solubility of $Ag_3PO_4$ in water is, $1.8\times 10^{-5}mol/L$