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Alexandra [31]
2 years ago
13

A key process common to all organisms is the ability to reproduce. Where do all cells come from?

Chemistry
1 answer:
satela [25.4K]2 years ago
5 0

Answer:

It’s d

Explanation:

It’s d 100%

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In your opinion, will all the objects in the Universe last forever? Justify your answer.​
Alona [7]

Answer:

Maybe (yes/no)

Explanation:

Cause some theories about the end of the universe. The fate of the universe is determined by its density. The preponderance of evidence to date, based on measurements of the rate of expansion and the mass density, favors a universe that will continue to expand indefinitely, resulting in the "Big Freeze" scenario below.

6 0
2 years ago
Convert the following: 4.6L(liters) to ml(milliliters).
Minchanka [31]

Answer:

D

Explanation:

6 0
2 years ago
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Given the incomplete equation representing a reaction:
Akimi4 [234]
O is what should go in the blank. O stands for Oxygen.
3 0
3 years ago
The cylinder of aluminum in class was approximately 13mm in circumference and 42mm in length. What would be it's approximate mas
Roman55 [17]

Answer:

Mass, m = 1.51 grams

Explanation:

It is given that,

The circumference of Aluminium cylinder, C = 13 mm = 1.3 cm

Length of the cylinder, h = 4.2 cm

We know that the density of the Aluminium is 2.7 g/cm³

Circumference, C = 2πr

r=\dfrac{C}{2\pi}\\\\r=\dfrac{1.3}{2\pi}\\\\r=0.206\ cm

Density is equal to mass per unit volume.

d=\dfrac{m}{V}

m is mass of the cylinder

V is the volume of the cylinder

V=\pi r^2h\\\\V=\dfrac{22}{7}\times0.206^2\times 4.2\\\\V=0.5601\ cm^3

So,

m=d\times V\\\\m=2.7\times 0.5601\\\\m=1.51\ g

So, the mass of the cylinder is 1.51 grams.

5 0
3 years ago
Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under s
Zarrin [17]

Answer:

90.3 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 NO(g) + O₂(g) → 2 NO₂(g)  ∆H°rxn = –114.2 kJ

We can find the standard enthalpy of formation for NO using the following expression.

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))

ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol

ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol

ΔH°f(NO(g)) = 90.3 kJ/mol

8 0
3 years ago
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