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kozerog [31]
3 years ago
7

Why can the same earthquake have different ratings on the 2 scales.

Chemistry
1 answer:
horsena [70]3 years ago
5 0
The scales are different
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August kekule described the various ring structures of the carbon compound benzene. What type of chemist would he be considered
Andreyy89
B.  He would be considered an Organic Chemist since Organic Chemistry is the study of Carbon and its compounds.
3 0
3 years ago
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A 16.0 mL sample of a 1.04 M potassium sulfate solution is mixed with 14.3 mL of a 0.880 M barium nitrate solution and this prec
stiks02 [169]
Number of moles in the K2SO4 sample
= (16/1000)*1.04= 0.01664 mol

Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol

Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.

The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%

Answer: 
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%

I apologize for the mistake previous to this update.

5 0
3 years ago
How many molecules are in 0.500 mole of N2O5?
Oksi-84 [34.3K]

Answer:

3,011.10e23.

Explanation:

3 0
3 years ago
What is the wavelength of radiation emitted when an electron goes from the n = 7 to the n = 4 level of the Bohr hydrogen atom? G
Phantasy [73]

Answer:

the wavelength of radiation emitted  is \mathbf{\lambda= 2169.62 \ nm}

Explanation:

The energy of the Bohr's hydrogen atom can be expressed with the formula:

\mathtt{E_n =- \dfrac{13.6\ ev}{n^2}}

For n = 7:

\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}

\mathtt{E_7 =-0.27755 \ eV}

For n = 4

\mathtt{E_4=- \dfrac{13.6\ ev}{4^2}}

\mathtt{E_4 =- 0.85\ eV}

The  electron goes from the n = 7 to the n = 4, then :

\mathtt{E_7-E_4 = (-0.27755 - (-0.85) ) \ eV}

\mathtt{= 0.57245\ eV}

Wavelength of the radiation emitted:

\mathtt{\lambda= \dfrac{hc}{0.57245 \ eV}}

where;

hc  = 1242 eV.nm

\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}

\mathbf{\lambda= 2169.62 \ nm}

4 0
3 years ago
Two students performed the same experiment separately and each one of them recorded two readings of mass which are given below.
Lapatulllka [165]

Answer:

Hey weirdo what's up?

So you got a question huh?

Lemme answer

As said the correct reading is 3.0grams

And the option A has 3.01 and 2.99 which are very mush precise and accurate to 3.0 gram don't you think?

So the answer is

Option ii the result of students A is both precise and accurate

Loye ya

Peace out

5 0
2 years ago
Read 2 more answers
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