1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Drupady [299]
3 years ago
5

How many moles is in 2.52*10^24 molecules of water?

Chemistry
1 answer:
nataly862011 [7]3 years ago
3 0
Hi friend
--------------
Your answer
-------------------

Water = H2O

Number of molecules in one mole of water = 6.022 × 10²³ [Avogadro's constant]

Given number of molecules = 2.52 × 10²³

So,
------

Number of moles =
\frac{2.52 \times 10 {}^{24} }{6.022 \times 10 {}^{23} }  \\  \\  = 4.184 \: (approximately)

HOPE IT HELPS
You might be interested in
Here is a list of ingredients for a simple cake:
cupoosta [38]
So really you only need to answer the first one because you have more than enough og everything. 2×(what)=14 then that will be your answer
3 0
4 years ago
What is the ph of a 0.027 M KOH solution?
Monica [59]
First you calculate the concentration of [OH⁻] in <span>solution :

POH  = - log [ OH</span>⁻]

POH = - log [ 0.027 ]

POH = 1.56

PH + POH = 14

PH + 1.56 = 14

PH = 14 - 1.56

PH = 12.44

hope this helps!

8 0
3 years ago
What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su
anastassius [24]

Answer:

V=27992L=28.00m^3

Explanation:

Hello,

In this case, the combustion of methane is shown below:

CH_4+2O_2\rightarrow CO_2+2H_2O

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3

Best regards.

8 0
3 years ago
3. Sodium carbonate (Na2CO3) is formed from the reaction between
user100 [1]

Na2C03 is formed through the reaction of

NaOH and H2CO3 namely sodium hydroxide and carbonic acid

NaOH -> strong base

H2CO3-> weak acid

8 0
3 years ago
Read 2 more answers
The following are products in the chemical reaction shown below.
scoundrel [369]

Answer:

C₆H₁₂O₆ and O₂

Explanation:

To know which option is correct, it important that we know the concepts of chemical equation .

A chemical equation gives us an expression of a chemical reaction in terms of symbols and formula. There are two sides to every equation. The side bearing the reactant and the side bearing the product.

The reactant is located on the left side and the product is located on the right side.

Reactant —> Product

With the above information, let us answer the question given above.

6CO₂ + 6H₂O —> C₆H₁₂O₆ + 6O₂

The products are located on the right side of the equation.

Thus, the products of the reaction is C₆H₁₂O₆ and O₂

8 0
3 years ago
Other questions:
  • If the atomic weight of nitrogen is 14.01, what is the mass of the nitrogen atoms in one mole of cadmium nitrate, Cd(NOÀ)
    8·1 answer
  • Sixty-five percent of the mass of bone is a compound called hydroxyapatite. sixty-five percent of the mass of bone is a compound
    15·1 answer
  • A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,
    5·1 answer
  • How many CO2 molecules in 0.0189 mol of CO2?
    5·1 answer
  • Bacteria, fungi and earthworms are all-​
    15·2 answers
  • Which of the following is an anion of 14N?<br> 14N5+<br> 14N3-<br> 15N
    5·1 answer
  • Match “the same” or “different” to each sentence.
    7·1 answer
  • How many atoms are in .938 moles of CO2?
    10·1 answer
  • According to Le
    14·2 answers
  • Hurry I need the answer asap
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!