How <br> many moles are in 8.5 •10^25

Carbonic acid is a mixture of

kirill115 [55]

Answer:

a compound of the elements hydrogen, carbon, and oxygen. (also carbonic acid is a compound not a mixture ^-^)

Explanation:

hope this helps dude

7 0

3 years ago

Read 2 more answers

You want to compare the malleability of

Paul [167]

One way you could measure this is hitting the two metals with a mallet and seeing which one has more of a dent

6 0

3 years ago

Liquid n-pentane and liquid n-octane are mixed to form a stream flowing at a rate of 1,496.5 lbm/hr. An in-line density measurem

Ivenika [448]

Answer:

202.20 cm^3/s

Explanation:

Hello,

I'm sending a photo showing the solution for this exercise.

The solution of the 2x2 system could be achieved by using a calculator or any available algebraic method.

Best regards.

4 0

3 years ago

An arctic weather balloon is filled with 4.96 L of helium gas inside a prep shed. The temperature inside the shed is 6. °C. The

Virty [35]

The new volume of the balloon : V₂ = 4.89 L

<h3>Further explanation</h3>

Given

V₁ = 4.96 L

T₁ = 6 + 273 = 279 K

T₂ = 2 + 273 = 275 K

Required

The new volume

Solution

Charles's Law

When the gas pressure is kept constant, the gas volume is proportional to the temperature

$\tt \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

Input the value :

V₂=V₁.T₂/T₁

V₂ = 4.96 x 275 / 279

V₂ = 4.89 L

3 0

3 years ago

A solution of fructose, C6H12O6, a sugar found in many fruits, is made by dissolving 34.0 g of fructose in 1.00 kg of water. Wha

Mashcka [7]

Answer:

Molality → 0.188 m

Mole fraction of fructose → 0.00337

Mass percent of fructose in solution → 3.29 %

Molarity → 0.183 M

Explanation:

Solute → 34 g of fructose

Solvent → 1000 g of water

Solution → 1000 g of water + 34 g of fructose = 1034 g of solution.

We take account density to calculate, the solution's density

1.0078 g/mL = 1034 g / mL

1034 g / 1.0078 g/mL = 1026 mL

Molal concentration → moles of solute in 1kg of solvent

Moles of fructose = mass of fructose / molar mass

34 g/ 180g/mol = 0.188 mol

0.188 mol/1kg = 0.188 m

Mole fraction of fructose = Moles of fructose / Total moles

We determine the moles of water

Moles of water = 1000 g / 18 g = 55.5 mol

Total moles = moles of fructose + moles of water

0.188 mol + 55.5 mol = 55.743 mol

0.188 mol / 55.743 mol = 0.00337

Mass percent = mass of fructose in 100 g of solution

(Mass of fructose / Total mass ) . 100 = (34 g /1034 g) . 100 = 3.29 %

Molarity = Moles of solute in 1L of solution

We can also say mmol of solute in 1 mL of solution

0.188 mol of fructose = 188 mmol of fructose

Molarity = 188 mmol / 1026 mL of solution = 0.183 M

8 0

3 years ago

How many moles are in 8.5 •10^25