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pickupchik [31]
2 years ago
13

Can someone help with this problem

Chemistry
1 answer:
gavmur [86]2 years ago
3 0
1. Fe2O3
2. Molar mass of Fe2O3 - (55.8x2)+(16x3) = 159.6
Moles =mass/Molar mass
Moles=9.2/159.6=0.058mol
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How much concentrated 18M sulfuric acid is needed to prepare 250mL of a 6.0M solution?
slavikrds [6]
C₁ * V₁ = C₂ * V₂ 

18 * V₁ = 6.0 * 250

18 V₁ = 1500

V₁ = 1500 / 18

V₁ = 83.33 mL

hope this helps!


5 0
3 years ago
Calculate the pH and pOH of 0.010 mol OH- per liter
Sati [7]
We can use a variety of formulas to determine our answers here.

Our formula for pOH is -log(mol), and we can plug it in as -log(0.010). Take note that OH- is a base, not an acid. 

So, the pOH of OH- is 2.

To find pH we can set up this simple equation:
pH + pOH = 14
All we need to do is subtract 2 from 14, therefore the pH is 12.

This makes sense since acids range in the pH of 1-6, and we are dealing with a base. Hope I could help!
7 0
2 years ago
A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
How does a filter separate mixtures like sand and water?​
Talja [164]

Answer:

Filtration is a method for separating an insoluble solid from a liquid. When a mixture of sand and water is filtered: the sand stays behind in the filter paper (it becomes the residue ) the water passes through the filter paper (it becomes the filtrate )

Explanation:

8 0
2 years ago
A sample of gas occupies 300 mL at 100K. What is its volume when the
iris [78.8K]

If the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.

<h3>What is Charles's law?</h3>

Charles's law states that "the volume occupied by a definite quantity of gas is directly proportional to its absolute temperature.

It is expressed as;

V₁/T₁ = V₂/T₂

Given the data in the question;

  • Initial temperature of gas T₁ = 100K
  • Initial volume of gas V₁ = 300mL
  • Final temperature T₂ = 200K
  • Final volume V₂ = ?

V₁/T₁ = V₂/T₂

V₂ = V₁T₂ / T₁

V₂ = ( 300mL × 200K ) / 100K

V₂ = 60000mLK / 100K

V₂ = 600mL

Therefore, if the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.

Learn more about Charles's law here: brainly.com/question/12835309

#SPJ1

3 0
2 years ago
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