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kramer
3 years ago
8

Two of the simplest compounds containing just carbon and hydrogen are methane and ethane. Methane contains 0.3357 g of hydrogen

for every 1.00 g of carbon. The ratio of grams of hydrogen per gram of carbon in methane to grams of hydrogen per gram of carbon in ethane is 4:3. Determine the number of grams of hydrogen per gram of carbon in ethane.
Chemistry
1 answer:
Andreyy893 years ago
5 0
Let the ratio of grams of hydrogen per gram of carbon in methane be M, we know that:
M = 0.3357 g / 1 g

Next, lets represent the grams of hydrogen per gram of carbon in ethane be E. The final piece of information we have is:

M / E = 4/3

If we cross multiply,

3M = 4E

Now, substituting the value of M from earlier and solving for E,

E = (3 * 0.3357) / 4
E = 0.2518

There are 0.2518 grams of hydrogen per gram of carbon in ethane.
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Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at
valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100  - x  

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,  

121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042

120.9038x+122.9042\left(100-x\right)=12176.01

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 %  = 42.8 %</u>

6 0
3 years ago
A metal ion Mⁿ⁺ has a single electron. The highest energy line in its emission spectrum occurs at a frequency of 2.961 x 10¹⁶ Hz
Whitepunk [10]

Answer:

z≅3

Atomic number is 3, So ion is Lithium ion (Li^+)

Explanation:

First of all

v=f*λ

In our case v=c

c=f*λ

λ=c/f

where:

c is the speed of light

f is the frequency

\lambda=\frac{3*10^8}{2.961*10^{16}}\\ \lambda=1.01317*10^{-8} m

Using Rydberg's Formula:

\frac{1}{\lambda}=R*z^2*(\frac{1}{n_1^2}-\frac{1}{n_2^2})

Where:

R is Rydberg constant=1.097*10^7

z is atomic Number

For highest Energy:

n_1=1

n_2=∞

\frac{1}{1.01317*10^{-8}}=1.097*10^{7}*z^2*(\frac{1}{1^2}-\frac{1}{\inf})\\z^2=8.99\\z=2.99

z≅3

Atomic number is 3, So ion is Lithium ion (Li^+)

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What modifications were made to the original Ray Cathode experiment?
Mariulka [41]
They added tubes thats the answer

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sesenic [268]

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electrophilic addition

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How many grams of NH3 can be produced from 4.98 mol of N2 and excess H2
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169.32 grams of NH3 ........
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