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Zanzabum
2 years ago
15

If the absorbance of the 10μM solution is 0.24, what should be the absorbance of the 50μM? Write answer to two decimal places.

Chemistry
1 answer:
Phoenix [80]2 years ago
3 0
Absorbance/Concentration = Constant.

That is, absorbance is directly proportional to concentration.

Therefore,

0.24/10 = x/50, where x is the absorbance at 50 micro m.

Calculating for x;
x=50* (0.24/10) = 1.20 (correct to 2 decimal places)
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write the net ionic equation (including phases) that corresponds to Cu(ClO4)2(aq) + Na2CO3(aq)-------->CaCO3(s) + 2NaClO4(aq)
soldier1979 [14.2K]
Complete ionic:
Cu(aq) + 2Cl(aq) + 8O(aq) + 2Na(aq) + C(aq) + 3O(aq) = CaCO3(s) + 2Na(aq) + Cl(aq) + 4O(aq) 

Net ionic:

Cu(aq) + Cl(aq) + 4O(aq) + 2Na(aq) + C(aq) + 3O(aq) = CaCO3(s) 

So write everything out as IF it will dissociate in water. So everything that is aq splits but solid just floats to the bottom of the mixture. Cancel what you can (in this case the two from the ClO4 on the left of the equation cancels with the ClO4 from the right) and the 2Na cancels. Then, write out the whole solution and you are done! 
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Pick the correct model
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8 0
3 years ago
1. Compare masses: a) 0,4mol CO₂ and 0,6mol H₂O ; b) 0,135mol H₂SO₄ and 0,5mol HCI.
bixtya [17]

Answer:

you can now deduct which one is greater or smaller and by how much.

Explanation:

no of moles= mass/molar mass

1ai) 0.4 = m/ ( 12 + (16*2)

m= 0.4* 44

m= 17.6g

ii) 0.6= m/( 2*1 + 16)

m= 0.6 *18

m= 10.8g

b) 0.135 = m/ ( 2*1 +32 + (16*4)

m= 0.135* 98

m= 13.23g

ii) 0.5= m/ (1+35.5)

m= 0.5*36.5

m= 18.25g

2. Avogadro's Number = 6.02×10²³

1 mol of any element= 6.02×10²³ particles

a) 0.1 mol of H20= (6.02×10²³) * 0.1

= 6.02×10²² molecules

ii) 0.3 mol of CO2= (6.02×10²³) * 0.3

= 1.806 × 10²³ molecules

Ans: 0.3 mol of CO2

bi) 0.25 mol of HCl= (6.02×10²³) * 0.25

= 1.505 × 10²³ molecules

bii) - find the no of moles first:

no of moles= mass/molar mass

n= 3.4g/ 34g →mr of H2S in g=2+32= 34g

n= 0.l mol

- use the Avogadro Number.

0.1 mol of H20= (6.02×10²³) * 0.1

= 6.02×10²² molecules.

biii) here you're given the density, use it to find the mass of acetic acid.

ρ = 1049 g/ml

ρ = m/v, where v=5 ml

1049 = m/ 5

m= ρ*v

m= 1049*5

m= 5245 g

• convert this into moles.

mr of CH3COOH= 12 + 3+ 12+ 16+ 16+ 1

= 60

mr in g = 60g

n= m/mr

n= 5245/ 60

n= 87. 41666...

n= 87.4 moles

•using Avogadro's Number:

87.4 moles of acetic acid=(6.02×10²³)*87.4

= 2.25148* 10²⁵

= 2.25 * 10²⁵ molecules

thus, the ans for this is 5 ml of acetic acid.

7 0
2 years ago
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