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Levart [38]
3 years ago
8

What if it is did Darwin used to support his theory of evolution

Chemistry
1 answer:
Tatiana [17]3 years ago
8 0

Darwin used finches in his theory of evolution.

You might be interested in
The concentration of copper(II) sulfate in one brand of soluble plant fertilizer is 0.070% by weight. If a 21.5 g sample of this
Semmy [17]

Answer:

The molar concentration of Cu^{2+} ions in the given amount of sample is 4.73\times 10^{-5}M

Explanation:

Given that,

Mass of sample = 21.5 g

0.07 % (m/m) of copper (II) sulfate in plant fertilizer

This means that, in 100 g of plant fertilizer, 0.07 g of copper (II) sulfate is present

So, in 20 g of plant fertilizer

,=\frac{0.07}{100}\times 21.5}\\=0.0151g of copper (II) sulfate is present.

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Mass of solute (copper (II) sulfate) = 0.0151 g

Molar mass of copper (II) sulfate = 159.6 g/mol

Volume of solution = 2.0 L

\text{Molarity of solution}=\frac{0.0151g}{159.6g/mol\times 2.0L}\\\\\text{Molarity of solution}=4.73\times 10^{-5}M

The chemical equation for the ionization of copper (II) sulfate follows:

CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}

1 mole of copper (II) sulfate produces 1 mole of copper (II) ions and sulfate ions

Molarity of copper (II) ions = 4.73\times 10^{-5}M

Hence, the molar concentration of Cu^{2+} ions in the given amount of sample is 4.73\times 10^{-5}M

4 0
3 years ago
What mass of oxygen (O2) forms in a reaction that forms 15.90 g C6H12O6? (Molar mass of O2 = 32.00 g/mol; molar mass of C6H12O6
Dafna11 [192]

The answer is: the mass of oxygen is 16.95 grams.

The overall balanced photosynthesis reaction:  

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.  

m(C₆H₁₂O₆) = 15.90 g; mass of glucose.

n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆).

n(C₆H₁₂O₆) = 15.9 g ÷ 180.18 g/mol.

n(C₆H₁₂O₆) = 0.088 mol; amount of glucose.

From chemical reaction: n(C₆H₁₂O₆) : n(O₂) = 1 : 6.

n(O₂) = 6 · 0.088 mol.

n(O₂) = 0.53 mol; amount of oxygen.

m(O₂) = 0.53 mol · 32.00 g/mol.

m(O₂) = 16.95 g; mass of oxygen.

5 0
3 years ago
Read 2 more answers
When a redox reaction that takes place in an acidic solution involves an oxygen imbalance, oxygen should be balanced by adding _
olya-2409 [2.1K]

Balancing redox reactions:

Oxygen should be balanced by adding H_{2}O  as needed, while hydrogen should be balanced by adding H^{+}.

What is a redox reaction?

Redox reactions, also known as oxidation-reduction reactions, involve the simultaneous oxidation and reduction of two different reactants.

The Half-Equation Method is one technique used to balance redox processes. The equation is divided into two half-equations using this technique: one for oxidation and one for reduction.

By changing the coefficients and adding H_{2}O, H^{+}, and e^{-} in that order, each reaction is brought into equilibrium:

  1. By putting the right number of water (H_{2}O) molecules on the other side of the equation, the oxygen atoms are brought into balance.
  2. By adding H^{+} ions to the opposing side of the equation, one can balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom).
  3. Total the fees for each side. Add enough electrons (e^{-}) to the more positive side to make them equal. (As a general rule, e^{-} and  H^{+}are nearly always on the same side.)
  4. The e^{-} on either side must be made equal; if not, they must be multiplied by the lowest common multiple (LCM) in order to make them equal.
  5. One balanced equation is created by adding the two half-equations and canceling out the electrons. Additionally, common terms should be eliminated.
  6. Now that the equation has been verified, it can be balanced.

Learn more about redox reaction here,

brainly.com/question/20068208

#SPJ4

7 0
2 years ago
(c) Assume you have an equilibrium mixture of [A], [B], and [C] at 298K and that the
djyliett [7]

Answer:

Explanation:

1. The amount of CaCO3 must be so small that  

P

CO

2

 is less than KP when the CaCO3 has completely decomposed. In other words, the starting amount of CaCO3 cannot completely generate the full  

P

CO

2

 required for equilibrium.

3. The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side.

5. No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.

7. Add N2; add H2; decrease the container volume; heat the mixture.

9. (a) ΔT increase = shift right, ΔP increase = shift left; (b) ΔT increase = shift right, ΔP increase = no effect; (c) ΔT increase = shift left, ΔP increase = shift left; (d) ΔT increase = shift left, ΔP increase = shift right.

11. (a)  

K

c

=

[

CH

3

OH

]

[

H

2

]

2

[

CO

]

; (b) [H2] increases, [CO] decreases, [CH3OH] increases; (c), [H2] increases, [CO] decreases, [CH3OH] decreases; (d), [H2] increases, [CO] increases, [CH3OH] increases; (e), [H2] increases, [CO] increases, [CH3OH] decreases; (f), no changes.

13. (a)  

K

c

=

[

CO

]

[

H

2

]

[

H

2

O

]

; (b) [H2O] no change, [CO] no change, [H2] no change; (c) [H2O] decreases, [CO] decreases, [H2] decreases; (d) [H2O] increases, [CO] increases, [H2] decreases; (f) [H2O] decreases, [CO] increases, [H2] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.

15. Only (b)

17. Add NaCl or some other salt that produces Cl− to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).

19. (a)

Hope this helps :)

3 0
3 years ago
What mass of cucl2 is contained in 75.85 g of a 22.4% by mass solution of cucl2 in water?
Sindrei [870]
First  step: convert   22.4%  into   fraction
 that  is  22.4/100  which  is equivalent  to 0.224
  The  mass  of Cucl2  contained  in  75.85  of  22.4%  by  mass  of   solution  of  CUcl2 in  water is   therefore,
0.224 x75.85=16.99g
alternatively
75.85 --->100%
      ?        22.4%
by cross  multiplication =(22.4 x 75.85)/100 =16.99g
5 0
3 years ago
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