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uysha [10]
2 years ago
14

Lana is testing her hypothesis that marigolds grow faster in red light than in yellow light. If the plants in yellow light grow

faster during her experiment what should Lana do next? A. Assume that she made a mistake during the experiment B. Report that her hypothesis was useless C. Conclude that the experiment did not work D. Repeat the experiment to confirm the result​
Chemistry
1 answer:
Alinara [238K]2 years ago
7 0

Answer:

in particular, chlorophyll absorbs blue and red light while allowing green light to be reflected (or transmitted). ... If chlorophyll needs red and blue light, what do you think would happen to the plant if you were to place a green filter over your light source so that ... Think of the rainbow, what colors are there other than green?

Explanation:

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At a temperature of __________ °c, 0.444 mol of co gas occupies 11.8 l at 889 torr.
Novay_Z [31]
<span>ideal gas law is: PV = nRT P = pressure (torr) = 889 torr V = volume (Liters) = 11.8 L n = moles of gas = 0.444 mol R = gas constant = 62.4 (L * torr / mol * k) solve for T (in kelvin) T = PV/nR T = (889*11.8)/(.444*62.4) T = 378.6 K convert to C (subtract 273) T = 105.6 deg C</span>
3 0
2 years ago
Read 2 more answers
(ii) When shale gas is burned, the hydrogen sulfide reacts with oxygen.
Elenna [48]

Answer:

See explanation

Explanation:

Now , we have the equation of the reaction as;

2H2S(g) + 302(g)------->2SO2(g) + 2H2O(g)

This equation shows that SO2 gas is produced in the process. Let us recall that this same SO2 gas is the anhydride of H2SO4. This means that it can dissolve in water to form H2SO4

So, when SO2 dissolve in rain droplets, then H2SO4 is formed thereby lowering the pH of rain water. This is acid rain.

4 0
2 years ago
Using complete subshell notation (not abbreviations, 1s 22s 22p 6 , and so forth), predict the electron configuration of each of
dybincka [34]

<u>Answer:</u> The electronic configuration of the elements are written below.

<u>Explanation:</u>

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

For the given options:

  • <u>Option a:</u>  Carbon (C)

Carbon is the 6th element of the periodic table. The number of electrons in carbon atom are 6.

The electronic configuration of carbon is 1s^22s^22p^2

  • <u>Option b:</u>  Phosphorus (P)

Phosphorus is the 15th element of the periodic table. The number of electrons in phosphorus atom are 15.

The electronic configuration of phosphorus is 1s^22s^22p^63s^23p^3

  • <u>Option c:</u>  Vanadium (V)

Vanadium is the 23rd element of the periodic table. The number of electrons in vanadium atom are 23.

The electronic configuration of vanadium is 1s^22s^22p^63s^23p^64s^23d^3

  • <u>Option d:</u>  Antimony (Sb)

Antimony is the 51st element of the periodic table. The number of electrons in antimony atom are 51.

The electronic configuration of antimony is 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^3

  • <u>Option e:</u>  Samarium (Sm)

Samarium is the 62nd element of the periodic table. The number of electrons in samarium atom are 62.

The electronic configuration of samarium is 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^6

Hence, the electronic configuration of the elements are written above.

4 0
2 years ago
what mass of lithium chloride, licl must be dissolved to make a 0.194M solution that has volume of 1.00 l
Bogdan [553]
MXV= (0.194M)(1.00L)=0.194moles
42.39LiClg/molex0.194moles=8.2g LiCl
6 0
2 years ago
Gold has always been a highly prized metal, and it has been widely used from the beginning of history as a store of value.It doe
Paul [167]

The equation given is incorrect, the correct equation is:

Au(s) + HNO₃(aq) + HCl(aq) → HAuCl₄(aq) + NO₂(g) + H₂O(l)

Answer:

B) It supplies chloride ions to form a complex ion with the oxidized gold.

Explanation:

The equation given is:

Au(s) + HNO₃(aq) + HCl(aq) → HAuCl₄(aq) + NO₂(g) + H₂O(l)

To balance the equation, we need to find the oxidation number (Nox) of the elements, and identify which substance is being oxided and which on is being reduced. Simple elements has Nox = 0, H has Nox = +1, O has Nox = -2. So:

Au(s): Nox Au = 0

HNO₃(aq): Nox H = +1; Nox O = -2, Nox N: +1 +x +3*(-2) = 0 -> x = +5

HCl(aq): Nox H = +1, Nox Cl = -1

HAuCl₄(aq): Nox H = +1, Nox Cl = -1, Nox Au: +1 +x +4*(-1) = 0 -> x = +3

NO₂(g): Nox O = -2; Nox N: x + 2*(-2) = 0 -> x = +4

H₂O(l): Nox H = +1; Nox O = -2

So, Au is being oxidezed from 0 to +3, and N is being reduced from +5 to +4:

ΔNox(Au) = 3

ΔNox(N) = 1

Both of them are single in the compound they are, so it's not necessary to multiply by the coefficient (because is 1). So, the ΔNox must be changed between the compounds:

Au(s) + 3HNO₃(aq) + HCl(aq) → HAuCl₄(aq) + NO₂(g) + H₂O(l)

Now, the balancing must be done by trial, knowing that the elements must be at the same number on both sides of the equation. So, we multiply HCl by 4, and NO₂ and H₂O by 3:

Au(s) + 3HNO₃(aq) + 4HCl(aq) → HAuCl₄(aq) + 3NO₂(g) + 3H₂O(l).

So, we can see that Au is being oxidized, so it's the reducing agent, and HNO₃ ins being reduced, so it's the oxidizing agent. The HCl supplies Cl ions to the complex formed with gold.

4 0
2 years ago
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