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OleMash [197]
2 years ago
8

What do animal and plant cells have in common

Chemistry
1 answer:
Pavlova-9 [17]2 years ago
6 0

Answer:

both have cell membranes

Explanation:

the animal cell has small  or temporary vacuoles.

animal cell does not have chloroplast or cell wall.

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A train that is accelerating at 5m/s² down the tracks and crashes into a car. If the train weighs 20,000kg, how much force does
iragen [17]

Answer:

100,000 N

Explanation:

Force, F = mass, m × acceleration, a

⇒F = ma

m= 20, 000kg

a = 5m/s²

∴ F = 5 × 20,000 = 100,000N

3 0
2 years ago
I know the photo trash but i need answers for the top middle and last question please :)
Firlakuza [10]

Answer:

Middle: Self heating containers are really expensive but are useful because they are easy to use, portable, and can be recycled, unlike a camp stove which is not as easily moved.

Explanation:

i neeed a better image for the first an last boo<3

6 0
2 years ago
Write a word equation for 2ZnS(s)+3O2(g) -&gt; 2ZnS(s) + 2SO3(g)
hjlf

2ZnS(s)+3O2(g) -> 2Zns(s) + 2SO3(g)

the above given equation is unbalanced as it contains 4 moles of sulphur in the output but in the input there are only two aoms of sulphur so to balance the equation we will write the equation as given under

balanced equation is

2ZnS(s)+3O2(g) -> 2Zn(s) + 2SO3(g)


In words:

When 2 moles of solid zinc sulfide reacts with 3 moles of oxygen gas gives 2 moles of solid zinc  and 2 moles of sulphur trioxide gas.

3 0
2 years ago
Some changes in matter may results to 1. _ or 2._ effects in the environment. We have to do our share in maintaining 3. environm
fredd [130]

Answer:

C

Explanation:

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>
6 0
2 years ago
The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate
Delicious77 [7]

<u>Answer:</u>

<u>For a:</u> The edge length of the crystal is 533.5 pm

<u>For b:</u> The atomic radius of potassium is 231.01 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the lattice parameter or edge length of the crystal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 0.855g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal = 39.09 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell = ?

Putting values in above equation, we get:

0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm

<u>Conversion factor:</u>  1cm=10^{10}pm

Hence, the edge length of the crystal is 533.5 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 533.5 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm

Hence, the atomic radius of potassium is 231.01 pm

4 0
3 years ago
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