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2 years ago

Can someone help me with my science it’s super hard :(

Chemistry

1 answer:

Lunna [17]2 years ago

5 0

Answer: here is a description of what they look like since I cannot see the diagram.

Mitochondria- wavy line

Ribosomes- little dots

Chromosome- X

Nucleus- Circle with bite taken out of it

Endoplasm- Bubble

Nuclear membrane- Outer layer/bubble

Golgi-wavy circle thing

Vacuole- Bigger circle (but not nucleus)

Cytoplasm- its the jelly on the inside

Cell wall- only in plant cells, looks like a box

Cell membrane- The outmost part of the animal cell, the barrier.

I sure hope this helps!!!

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If you combine 290.0 mL 290.0 mL of water at 25.00 ∘ C 25.00 ∘C and 140.0 mL 140.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t

Alexxandr [17]

Answer:

The final temperature is 47.79 °C

Explanation:

Step 1: Data given

Sample 1 has a volume of 290.0 mL

Temperature of sample 1 = 25.00 °C

Sample 2 has a volume of 140.00 mL

Temperature of sample 2 = 95.00 °C

Step 2: Calculate the final temperature

Heat lost = heat gained

Qlost = -Qgained

Q = m*c* ΔT

Q(sample1) = -Q(sample2)

m(sample1) * c(sample1) * ΔT(sample1) = -m(sample2)*c(sample2) *ΔT(sample2)

⇒with m(sample1) = the mass of sample 1 = 290.0 mL * 1g/mL = 290 grams

⇒with c(sample 1) = the specific heat of water = c(sample 2)

⇒with ΔT(sample 1) = the change of temperature = T2 - T1 = T2 - 25.00 °C

⇒with m(sample2) = the mass of sample 2 = 140.0 mL * 1g/mL = 140 grams

⇒with c(sample2) = the specific heat of water = c(sample1)

⇒with ΔT(sample2) = T2 -T1 = T2 - 95.00°C

m(sample1) * ΔT(sample1) = -m(sample2)*ΔT(sample2)

290 *(T2-25.0) = -140 *(T2 - 95.0)

290 T2 - 7250 = -140 T2 + 13300

430 T2 = 20550

T2 = 47.79 °C

The final temperature is 47.79 °C

4 0

3 years ago

The half-life of 3H (H-3) is 12 years. About how long does it take for 127/128 of a sample of that radionuclide to decay? (Hint:

Semenov [28]

Answer:

i do believe its 7

Explanation:

8 0

2 years ago

The solubility of BaCO3(s) in water at a certain temperature is 4.4 10–5 mol/L. Calculate the value of Ksp for BaCO3(s) at this

azamat

Answer : The value of $K_{sp}$ for $BaCO_3$ is $19.36\times 10^{-10}mole^2/L^2$ .

Solution : Given,

Solubility of $BaCO_3$ in water = $4.4\times 10^{-5}mole/L$

The barium carbonate is insoluble in water, that means when we are adding water then the result is the formation of an equilibrium reaction between the dissolved ions and undissolved solid.

The equilibrium equation is,

$BaCO_3\rightleftharpoons Ba^{2+}+CO^{2-}_3$

Initially - 0 0

At equilibrium - s s

The Solubility product will be equal to,

$K_{sp}=[Ba^{2+}][CO^{2-}_3]$

$K_{sp}=s\times s=s^2$

$[Ba^{2+}]=[CO^{2-}_3]=s=4.4\times 10^{-5}mole/L$

Now put all the given values in this expression, we get the value of solubility constant.

$K_{sp}=(4.4\times 10^{-5}mole/L)^2=19.36\times 10^{-10}mole^2/L^2$

Therefore, the value of $K_{sp}$ for $BaCO_3$ is $19.36\times 10^{-10}mole^2/L^2$ .

3 0

3 years ago

Lithium diorganocopper (Gilman) reagents are prepared by treatment of an organolithium compound with copper(I) iodide. Decide wh

Ann [662]

Answer:

See explanation and image attached

Explanation:

The Gilman reagent is a lithium and copper (diorganocopper) reagent with a general formula R2CuLi. R is an alkyl or aryl group.

They are useful in the synthesis of alkanes because they react with organic halides to replace the halide group with an R group.

In this particular instance, we intend to synthesize propylcyclohexane. The structure of the lithium diorganocopper (Gilman) reagent required is shown in the image attached to this answer.

7 0

3 years ago

How do you draw acetone using a Bohr model?

Inessa [10]

Explanation:

this is the one i found.....

6 0

3 years ago