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Yanka [14]
3 years ago
5

4,600,000,000,000 in scientific notation

Chemistry
1 answer:
BabaBlast [244]3 years ago
6 0
4.6to the twelth power plus ten or something like that it has been a while since I have done the scientific notation but where ever you put in the number of zeros. is twelve
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Which are forms of frozen water? Check all that apply. dew frost hail rain sleet
MariettaO [177]

When moist air cools below its dew point on a cold surface it forms dew. It is in the liquid state.

Rain is again a form of precipitation in which water is in the liquid state.

When air temperature is below freezing, the precipitation that results is referred to as frost. Hail is a form of precipitation in the ice balls whereas as sleet is a mixture of rain and snow.

Ans: Forms of frozen water- frost, hail and sleet

7 0
3 years ago
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Only the _____ word in an organism's complete scientific name has its first letter capitalized
dimaraw [331]
Only the first word in an organism's complete scientific name has its first letter capitalized.

When writing out the organism's scientific name, it goes Genus and then species. The Genus is always capitalized and species is always lowercase. Also when writing the name, usually you either write it in italics or underline the scientific name.
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4 years ago
What is formed by the hydrolysis of Starch?
Ymorist [56]

Answer:

soluble starch, maltose and various dextrins.

Explanation:

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3 years ago
For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the
Aneli [31]
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
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8 0
3 years ago
The Lyman series results from excited state hydrogen atoms transiting to
Nutka1998 [239]

Answer:

I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.

n = 5 4th excited state

n = 4 3rd excited state

n = 3 2nd excited state

n = 2 1st excited state

n = 1 ground state

Here are the possible spectral lines.

n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.

n = 4 to 3, 4 to 2, 4 to 1 = 3 lines

n = 3 to 2, 3 to 1 = 2 lines

n = 2 to 1 = 1 line. Add 'em up. I get 10.

b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.

c.The E for any level is -21.8E-19 Joules/n^2

To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.

So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.

Explanation:

8 0
3 years ago
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