Write short notes on different planet of the solar system

What is the instantaneous rate of formation of product C given the following information: a. Stoichiometric equation A+ B2C+ D b

quester [9]

Answer:

1.728 mol /(L*min)

Explanation:

Hello,

In the attached photo, you'll find the numerical procedure for your question.

- Take into account that the negative sign is eligible for reagents and positive for products.

Best regards!

6 0

3 years ago

Which of these would be easily soluble in water? a. salt b. oil c. aluminum foil

mote1985 [20]

A) Salt, because salt can easily dissolve in water. Oil would not dissolve or evaporate in water (think of an oil spill - does that oil dissappear?). Aluminum foil would definitely not dissolve in water, so it is not soluble.

5 0

3 years ago

Read 2 more answers

BEST GETS BRAINLIEST!!

tatuchka [14]

Gee. I'll have to guess at what's "commonly thought".

One thing is the scale. Nobody has an accurate picture of the scale in
his head, because we never see a true-scale drawing. THAT's because
it's almost impossible to draw one on paper.

Example:
Shrink the solar system and everything in it so that the Sun
is the size of a quarter (the 25¢ coin).
Then:
-- The Earth is in orbit around the sun, 8.6 feet from it.
That's close enough that you might think you could find the
shrunken Earth. Unfortunately, it's only 0.009 inch in diameter.

-- The shrunken Jupiter is a 'huge' gas giant almost 0.1 inch in diameter.
It's orbiting the sun, about 45 feet away from it.

-- The shrunken Uranus is another gas giant, about 0.035 inch in diameter.
It's orbiting the sun, about 165 feet away from it.

-- The nearest star outside of the solar system is 441 MILES away !
On the same shrunken scale !
And there's NOTHING between here and there !

I think that's the biggest point to make about the REAL solar system ...
its utter emptiness. With the sun reduced to something you can hold
in your hand, the planets are the size of grains of sand, with hundreds
of feet of nothingness between them.

Same for its mass: The solar system is approximately nothing but a star.
That's it. A star, with some dust and some gas around it, and here and there
in the neighborhood a microscopic pebble or a chip of mineral. But mostly
it's nothing but a star ... if you went around and gathered up all that other
rubbish in the same bag and called it a part of the same solar system, the
sun would still have more than 99% of the total mass, and the bag would
hold less than 1% of it.

Book ... It's getting late, Hillary's fading, and that's all I can think of.
I hope this much is some help.

3 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

A) Compute the repeat unit molecular weight of polystyrene. B) Compute the number-average molecular weight for a polystyrene for

Andrej [43]

Answer:

Explanation:

In Polystrene, the molecular formula for the repeat unit = $C_8H_8$ ;

and the atomic weights of Carbon C = 12.01 g/mol

For Hydrogen, it is 1.01 g/mol

Hence, the repeat unit molecular weight is:

m = 8 (12.01 g/mol)+8(1.01 g/mol)

m = 96.08 g/mol + 8.08 g/mol

m = 104.16 g/mol

The degree of polymerization = no-average molecular weight/repeat unit molecular weight.

Mathematically;

$DP = \dfrac{\overline M_n}{m}$

$\overline M_n= DP \times m$

$\overline M_n= 25000 \times 104.16 \ g/mol$

$\overline M_n= 2604000 \ g/mol$

7 0

2 years ago

Write short notes on different planet of the solar system​

Write short notes on different planet of the solar system