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gulaghasi [49]
3 years ago
9

Which element would make a poor, brittle conductor? use the periodic table for help. a. sulfur b. boron c. cobalt d. calcium?

Chemistry
1 answer:
Greeley [361]3 years ago
8 0
I think the answer is C or B
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What three things happen in order for a chemical reaction to occur?
MissTica
Three things that should happen for a chemical reaction to occure :
1. Position of electrons must change
2. Chemical bonds must be formed and broken between atoms and
3. No changes should happen to the nuclei
4 0
3 years ago
the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal
Sunny_sXe [5.5K]

Rydberg formula is given by:

\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )

where, R_{H} = Rydberg  constant = 1.0973731568508 \times 10^{7} per metre

\lambda = wavelength

n_{1} and n_{2} are the level of transitions.

Now, for n_{1}= 2 and n_{2}= 6

\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )

= 1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )

= 1.0973731568508 \times 10^{7} \times (0.25-0.0278 )

= 1.0973731568508 \times 10^{7} \times 0.23

= 0.2523958\times 10^{7}

\lambda = \frac{1}{0.2523958\times 10^{7}}

= 3.9620\times 10^{-7} m

= 396.20\times 10^{-9} m

= 396.20 nm

Now, for n_{1}= 2 and n_{2}= 5

\frac{1}{\lambda} = 1.0973731568508 \times  10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )

= 1.0973731568508 \times 10^{7} \times (0.25-0.04 )

= 1.0973731568508 \times 10^{7} \times (0.21 )

= 0.230 \times  10^{7}

\lambda= \frac{1}{0.230 \times 10^{7}}

= 4.3478 \times 10^{-7} m

= 434.78\times 10^{-9} m

= 434.78 nm

Now, for n_{1}= 2 and n_{2}= 4

\frac{1}{\lambda} = 1.0973731568508 \times  10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )

=  1.0973731568508 \times 10^{7} \times (0.25-0.0625 )

= 1.0973731568508 \times 10^{7} \times (0.1875 )

= 0.20575 \times 10^{7}

\lambda= \frac{1}{0.20575 \times 10^{7}}

= 4.8602 \times 10^{-7} m

= 486.02 \times 10^{-9} m

= 486.02 nm

Now, for n_{1}= 2 and n_{2}= 3

\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )

=  1.0973731568508 \times 10^{7} \times (0.25-0.12 )

=  1.0973731568508 \times 10^{7} \times (0.13 )

= 0.1426585\times 10^{7}

\lambda= \frac{1}{0.1426585\times 10^{7}}

= 7.0097 \times 10^{-7} m

= 700.97 \times 10^{-9} m

= 700.97 nm



5 0
2 years ago
Read 2 more answers
The following data were measured for the reaction BF3(g)+NH3(g)→F3BNH3(g): Experiment [BF3](M) [NH3](M) Initial Rate (M/s) 1 0.2
Rama09 [41]

Answer:

-r_{A}=k\times[BF_3]^{1}\times[NH_3]^{1}

Explanation:

The rate law of a chemical reaction is given by

-r_{A}=k\times[BF_3]^{\alpha}\times[NH_3]^{\beta}

This law can be written for any experiment, and making the quotient between those expressions the reaction orders can be found

Between experiments 1 and 2  

\frac{-r_{A1}}{{-r}_{A2}}=\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)^\beta

Then the expression for the calculation of \beta

\beta=\frac{ln\frac{-r_{A1}}{-r_{A2}}}{ln\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)}=\frac{ln\frac{0.2130}{0.1065}}{ln\left(\frac{0.250}{0.125}\right)}

Resolving  

\beta=1

Doing the same between experiments 3 and 4 the expression for \alpha is

\alpha=\frac{ln\frac{-r_{A3}}{-r_{A4}}}{ln\left(\frac{\left[BF_3\right]_3}{\left[BF_3\right]_4}\right)}=\frac{ln\frac{0.0682}{0.1193}}{ln\left(\frac{0.200}{0.350}\right)}

Resolving  

\alpha=1

This means that the rate law for this reaction is  

-r_{A}=k\times[BF_3]^{1}\times[NH_3]^{1}

5 0
3 years ago
Electrons always fill orbitals in the same order. Each s orbital holds 2 electrons, each set of p orbitals holds 6 electrons, ea
arsen [322]
Oxygen is the answer :)

4 0
3 years ago
Read 2 more answers
Highest ionization energy: Li, Be, B
Ad libitum [116K]

Hello!


On the periodic table, as we go down the periodic table, the ionization energy decreases, but as we go across the periodic table (left to right), the ionization increases.


On the periodic table, lithium (Li) is located in column one, beryllium (Be) is located in column two, and (B) boron is located in column 13. As stated above, when we go across the periodic table (left to right), the ionization increases.


Therefore, the element with the highest ionization energy is Boron, or symbol B on the period table.

3 0
3 years ago
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