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
Nonamiya [84]
3 years ago
13

Find a mole of 0.0960 g of H2SO4

Chemistry
1 answer:
nignag [31]3 years ago
3 0
Molar mass H₂SO₄ = 98.079 g/mol

1 mol -------- 98.079 g
? mole ------ 0.0960 g

moles = 0.0960 * 1 / 98.079

= 0.0960 / 98.079

= 9.788 x 10⁻⁴ moles

hope this helps!
You might be interested in
Select all that apply. Which of the following are characteristics of bases. contain hydroxide ion or produce it in a solution ta
Blababa [14]
I know tasting bitter is one
3 0
3 years ago
Read 2 more answers
How many grams of kclo3 can be dissolved in 100 grams of water at 30 degrees Celsius
kondaur [170]
<span>The solubility of KClO</span>₃ : ( 10.1 / 100 g water ) at 30ºC

10.1 g ------------ 100 g ( H₂O )
    ? g ------------- 100 g ( H₂O )

Mass of KClO₃ :

100 * 10.1 / 100

1010 / 100 = 10.1 g of KClO₃

hope this helps!
8 0
3 years ago
Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an
Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g    [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0
3 years ago
A mixture of two gases has a total pressure of 5.7 atm . If one gas has a partial pressure of 4.1 atm , what if the partial pres
Mrrafil [7]
If the two gases has a total pressure of 5.7 atm and one of the gases has a partial pressure of 4.1 the the other one has the pressure of 1.6

6 0
3 years ago
Read 2 more answers
50.0 ml of 0.010m naoh was titrated with 0.50m hcl using a dropper pipet. if the average drop from the pipet has a volume of 0.0
creativ13 [48]

25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.

The equation of the reaction is;

NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)

We can use the titration formula;

CAVA/CBVB = NA/NB

CA= concentration of acid

VA = volume of acid

CB = concentration of base

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

CB = 0.010 M

VB = 50.0 ml

CA = 0.50 M

VA = ?

NA = 1

NB = 1

Substituting values;

CAVANB = CBVBNA

VA =  0.010 ×  50.0 × 1/ 0.50 × 1

VA = 1 ml

Since the total volume of acid used is 1 ml and each drop contains 0.040 ml

The number of drops required is 1ml/0.040 ml = 25 drops

Learn more: brainly.com/question/1527403

4 0
3 years ago
Other questions:
  • Which adaptation would be most likely to help a prey species avoid predators?
    13·2 answers
  • The degree to which a measurement approaches a standard value is _____. accuracy
    14·2 answers
  • Express 0.08010 in exponential notation to the proper significant figures
    11·1 answer
  • What is an example of how to supply activation energy to begin reaction
    11·1 answer
  • How do I solve this problem? What is the correct answer?
    12·1 answer
  • A car has a displacement of 160 kilometers to the north in 2 hours. What is it’s velocity in kilometers per hour
    11·1 answer
  • Give two examples of good conductors of electricity
    15·1 answer
  • How many grams of CO2 would be made from 3.0 g of C6H6? Please no links
    13·1 answer
  • NEED QUICK ANSWERS ( science)
    7·1 answer
  • A sample of gas occupies 300 mL at 100K. What is its volume when the
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!