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saul85 [17]
2 years ago
3

Which of the following statements best describes the process of photosynthesis?

Chemistry
1 answer:
Luden [163]2 years ago
7 0

Answer:

i dont see any follwing statements but photosynthesis is

Explanation:

Photosynthesis, the process by which green plants and certain other organisms transform light energy into chemical energy. During photosynthesis in green plants, light energy is captured and used to convert water, carbon dioxide, and minerals into oxygen and energy-rich organic compounds.

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What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c
brilliants [131]

Answer:

Molecular formula → PbSO₄ → Lead sulfate

Option c.

Explanation:

The % percent composition indicates that in 100 g of compound we have:

68.3 g of Pb, 10.6 g of S and  (100 - 68.3 - 10.6) = 21.1 g of O

We divide each element by the molar mass:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

We divide each mol by the lowest value to determine, the molecular formula

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Molecular formula → PbSO₄ → Lead sulfate

8 0
2 years ago
Read 2 more answers
Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle
Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

6 0
2 years ago
What is the freezing point of a 1.40 m aqueous solution of a nonvolatile un-ionized solute? (the freezing point depression const
Alina [70]
Depression is freezing point is a colligative property. It is mathematically expressed as ΔTf = Kf X m

where Kf = <span>freezing point depression constant = 1.86°c kg /mol (for water)
m = molality of solution = 1.40 m

</span>∴ ΔTf = Kf X m  = 1.86 X 1.40 = 2.604 oC

Now, for water freezing point = 0 oC

∴Freezing point of solution = -2.604 oC
6 0
2 years ago
Kayla owns a food truck that sells tacos and burritos. She sells each taco for $3 and each burrito for $7.25. Yesterday Kayla ma
Naddik [55]
So the put a lot of words to make this seem more complicated than it is. Your first equation involves the money. I’m going to use x to represent tacos and y to represent burritos.
3x+7.25y=595 would be your first equation because you know the price of each item already just not how many. The second would involve the twice as many burritos sold than tacos. So that would mean x+2 would equal y
Y=x+2
Hope this helps. If not I can explain it more in detail.
6 0
2 years ago
An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the
victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

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

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M

  • To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

pOH=14-8.08=5.92

  • To calculate pOH of the solution, we use the equation:

pOH=-\log[OH^-]

Putting values in above equation, we get:

5.92=-\log[OH^-]

[OH^-]=10^{-5.92}=1.202\times 10^{-6}M

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of X^- ions follows:

                    X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b

<u>Initial:</u>              0.1

<u>At eqllm:</u>        0.1-x                           x              x

Concentration of OH^-=x=1.202\times 10^{-6}M

The expression of K_b for above equation follows:

K_b=\frac{[OH^-][HX]}{[X^-]}

Putting values in above expression, we get:

K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M

  • To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:  

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant

K_b = Base dissociation constant = 1.445\times 10^{-11}

Putting values in above equation, we get:

10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}

We know that:

K_a\text{ for HF}=6.8\times 10^{-6}

K_a\text{ for HCl}=1.3\times 10^{6}

K_a\text{ for HClO}=3.0\times 10^{-8}

So, the calculated K_a is approximately equal to the K_a of HF

Hence, the unknown salt is NaF

6 0
2 years ago
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