All categories

3 years ago

Plz help ya' girl out! Label the parts of whatever that is Plz giving brainliest!

Chemistry

1 answer:

Burka [1]3 years ago

4 0

Answer:

BRAINLIEST PLEASE!!!

A = Smooth Endoplasmic Reticulum

B = Vacuole

C = Mitochondria (Plural)

D = Nucleolus

E = Cell Wall

Explanation:

This is obviously a plant cell. Animal cells do not have cell walls(E), and most animal cells will not have a vacuole(B). Smooth Endoplasmic Reticulum: This produces protiens (contains ribosomes) but when looked through a microscope, will appear to be smooth. The vacuole(B) is like a storage for the plant cell, and it helps support the plant cell. This is the reason why plants “droop” when you don’t water them for a long time. Mitochondia (Plural) processes nutrients for the cell. The nucleolus(D) covers the nucleus. It’s main function is to process RNA and combine it with proteins. The cell wall(E) helps support and protect the cell.

You might be interested in

How to calculate the frequency of an ion

sergij07 [2.7K]

solution:

You need to find the frequency, and they have already given you the wavelength. And since you already know the speed of light, you can use formula (2) to answer this problem. Remember to convert the nano meters to meters because the speed of light is in meters. $(1 nm = 1.0 x 10^{-9} m)$

$v=c\lambda \\v= (3.00 \times 10^8 m/s)/(6.9\times 10^-7 m) \\v = 4.35 \times 10^{14} s^{-1}$

8 0

3 years ago

The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?

Aloiza [94]

Answer:

0.0468 g.

Explanation:

The decay of radioactive elements obeys first-order kinetics.

For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.

For first-order reaction: kt = lna/(a-x).

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).

5.54688 = ln(12)/(a-x).

Taking e for the both sides:

256.34 = (12)/(a-x).

∴ (a-x) = 12/256.34 = 0.0468 g.

8 0

3 years ago

46. The number of orbitals in the sub-levels increases by even numbers. a. TRUE b. FALSE

frosja888 [35]

Answer:

It's true.

Explanation:

In sub-energy level : number of orbitals

s : 2
p : 6
d : 10
f : 14

$.$

6 0

3 years ago

What is the place of metalloids in the periodic table?

sleet_krkn [62]

B,Si,Ge,As,Sb,Te, and At are all the metalloids on the periodic table of elements.

3 0

3 years ago

A reactant decomposes with a half-life of 29.5 s when its initial concentration is 0.229 M. When the initial concentration is 0.

Dmitrij [34]

Answer :

The order of reaction is, 0 (zero order reaction).

The value of rate constant is, $0.00388Ms^{-1}$

Explanation :

Half life : It is defined as the time in which the concentration of a reactant is reduced to half of its original value.

The general expression of half-life for nth order is:

$t_{1/2}\propto \frac{1}{[A_o]^{n-1}}$

or,

$\frac{t_{1/2}_1}{t_{1/2}_2}=\frac{[A_2]^{n-1}}{[A_1]^{n-1}}$

or,

$n=\left(\frac{\log\frac{(t_{1/2})_1}{(t_{1/2})_2}}{\log\frac{(A)_2}{(A)_1}}\right )+1$ .............(1)

where,

$t_{1/2}$ = half-life of the reaction

n = order of reaction

[A] = concentration

As we are given:

Initial concentration of A = 0.229 M

Final concentration of A = 0.639 M

Initial half-life of the reaction = 29.5 s

Final half-life of the reaction = 82.3 s

Now put all the given values in the above formula 1, we get:

$n=\left (\frac{\log \frac{29.5}{82.3}}{\log\frac{0.639}{0.229}}\right )+1$

$n=0.000196\approx 0$

Thus, the order of reaction is, 0 (zero order reaction).

Now we have to determine the rate constant.

To calculate the rate constant for zero order the expression will be:

$t_{1/2}=\frac{[A_o]}{2k}$

When,

$t_{1/2}$ = 29.5 s

$[A_o]$ = 0.229 M

$29.5s=\frac{0.229M}{2k}$

$k=0.00388Ms^{-1}$

Thus, the value of rate constant is, $0.00388Ms^{-1}$

4 0

3 years ago