Mass of 2.88x10^-4 mol of ascorbic acid

Answers

Answer 1

Answer: Molar mass C₆H₈O₆ = 176.12 g/mol

1 mole ------------------ 176.12 g
2.88x10⁻⁴ mole ------- ?

mass = 2.88x10⁻⁴ * 176.12 / 1

mass ≈ 0.05072 g

hope this helps!

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The molar heat of fusion of gold is 12.550 kJ mol–1. At its melting point, how much mass of melted gold must solidify to release 235.0 kJ of energy?

Answers

The total heat released, Q, equals number of moles, n, times molar heat of fussion, Hf. this is: Q = n * Hf. You kno Q and Hf, so you can find n as n = Q/Hf = 235.0 kJ / 12.55 kJ/mol = 18.72 mol. Now you can pass that to mass using the atomic mass of Au, which is 197 g/mol => mass = 18.73 mol * 197 g/mol = 3688.9 grams.

Answer:

the answer is d.

Explanation:

Excuse me while I go kill Thanos.

10 divided by 2 im confused

Answers

if your serious about this question then it is 5

The awncer is 5!! Just use a Calculator and you would’ve gotten 5!!

Why does a high temperature have to be used to weld pieces of metal together

Answers

because of the metling point of metals, it is difficult to just simply put metal together and have them to stay together without any type of glue or epoxy based creams.

wielding is used to melt the surface or even change the compound completely so they are able to combine with other metals.

same thing with atoms, you need a lot of heat, and speed to simply make salt or even copper because of the elements not really being favorable of each other.

If an excess of sulfric acid reacts with 23.0 grams of sodium chloride how many grams of hydrochloric acid are produced? 2H2SO4 + 4NaCI -> 2H2SO4 + 4HCI

Answers

NaCl = 58.44 g/mol
H₂SO₄ = 98.0 g/mol

H₂SO₄ + 2 NaCl = 2 HCl + Na₂SO₄

98.0 g H₂SO₄ -----------> 2 x 58.44 g NaCl
? g H₂SO₄ ------------> 23.0 g NaCl

Mass of H₂SO₄ = 23.0 x 98.0 / 2 x 58.44

Mass of H₂SO₄ = 2254 / 116.88

=> 19.284 g of H₂SO₄

hope this helps!

I am in desperate need of help will give 40 points to whoever helps me!!!!! I have a quiz tomorrow but I have no idea how to do these practice problems. I know u have to use the boiling and freezing pint formula but i don't know how to plug them in because these problems all look different and asks for different things

Answers

All of the questions here are pertaining to the colligative properties of a solution and the preparation of solutions. Maybe, it would be best if you understand the equations to be used in order to answer these questions.
Freezing point depression or Boiling point elevation:

ΔT = -K (m) (i)

ΔT is the change in the freezing point or the boiling point not the freezing point/boiling point. Therefore, it should be added to the original value of the property of the solvent.

K is a constant called the molal freezing point depression constant and for the boiling point is the boiling point elevation constant. It is a property of the solvent.

m is the concentration of the solute in the solvent in terms of molality or kg solute/kg solvent.

i is the vant hoff factor which will represent the number of ions which the solute dissociates when in solution.

How many molecules are there in 79g of Fe2O3? how many atoms is thi

Answers

Molar masses:
$m_(Fe) = 56g/mol\nm_(O) = 16g/mol\n\n\therefore m_{Fe_(2)O_(3)} = 2 \cdot (56)+3 \cdot (16) = 112+48 = 160g/mol$

So:

$1mol \ Fe_(2)O_(3) = 160g = 6 \cdot 10^(23)molecules$

160g ---------- 6·10²³molecules
79g ---------- x

x = 2.9625·10²³ molecules

1molecule = 5 atoms
2.9625·10²³ molecules = 2.9625·10²³·5 = 14.8125·10²³ atoms

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