Ask question

Ask question

All categories

3 years ago

6

Is it better for an element to be more abundant or less abundant?

1 answer:

PIT_PIT [208]3 years ago

8 0

I believe it's better for it to be more abundant because it's better to have more then less. Take water as an example. Water is an important factor to our survival and if we didn't have enough of that we would be in trouble. Or if we needed food but there wasn't enough more people would be dying from starvation. It's better to have a little more than a little less.

You might be interested in

Why is earth likely unique in it volume of liquide water

Karolina [17]

It is unique because we can drink from it .

4 0

3 years ago

What MOSTLY determines the chemical properties of the atoms of an

Burka [1]

C......................

8 0

3 years ago

Consider a pot of water at 100 C. If it took 1,048,815 J of energy to vaporize the water and heat it to 135 C, how many grams of

jeka57 [31]

Answer:

There was 450.068g of water in the pot.

Explanation:

Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L

Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s

Let m = x g be the weight of water in the pot.

Energy required to vaporise water = mL = 2260x

Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x

Total energy required = $2260x+x\times2.010\times(135-100)=2260x+70.35x=2330.35x$

$2330.35x=1048815\\x=450.068g$

Hence, there was 450.068g of water in the pot.

8 0

3 years ago

Chlorine has an atomic number of 17.

zlopas [31]

Answer:

A. 17

Explanation:

6 0

4 years ago

Read 2 more answers

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

7 0

4 years ago