What are the nonmentals in group 16 of the periodic table?

**PLATO QUESTION, PLEASE ANSWER CORRECTLY, THANKS SO MUCH**

OLga [1]

None of the questions asked can be answered completely from the graph provided (GHG emissions: Direct, indirect and total Vs Year)

Reason:
1) Question A:What caused a drop in GHG emissions around 2009?. This questions in pointing towards reason for drop of GHG emission around 2009. From the graph, it can be seen that there is a drop in GHG emission around 2009. However, information for reason for this drop is not available in graph.
2) Question B: Did GHG emissions cause the melting of Arctic glaciers?. As mentioned earlier, the graph plotted provides information of GHG emissions: Vs Year. Information related to impact of GHG on environment is not available in graph.
3) Question C: How much methane was emitted by homes between 1990 and 2000?. Graph provides information of direct and indirect emission for GHG. However, it lacks information about emission from residential or industrial sources.
4) Question D: Does industrial equipment release gases other than greenhouse gases?: Present study doesnot cover type of gases emitted from industrial equipment.
5) Question E: Which types of industries were included in the study?: Present graph has not specific information related to industries.

6 0

3 years ago

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The reactant concentration in a first-order reaction was 8.10×10−2 M M after 15.0 s s and 1.80×10−3 M M after 90.0 s s . What is

kipiarov [429]

Answer:

The answer to the question is

The rate constant for the reaction is 1.056×10⁻³ M/s

Explanation:

To solve the question, e note that

For a zero order reaction, the rate law is given by

[A] = -k×t + [A]₀

This can be represented by the linear equation y = mx + c

Such that y = [A], m which is the gradient is = -k, and the intercept c = [A]₀

Therefore the rate constant k which is the gradient is given by

Gradient = $\frac{[A]_{2} - [A]_{1} }{t_{2} - t_{1} }$ where [A]₁ = 8.10×10⁻² M and [A]₂ = 1.80×10⁻³ M

= $\frac{1.80*10^{-3} M- 8.10*10^{-2} M}{90 s - 15 s}$ = -0.001056 M/s = -1.056×10⁻³ M/s

Threfore k = 1.056×10⁻³ M/s

3 0

3 years ago

Helllppppppppp Don’t Guess

ruslelena [56]

Answer:

C. Lithium

Explanation:

I Goo gled it and I think that's right.

3 0

2 years ago

Human blood consists of blood groups A, B, and O. However, there is one more blood group, AB, which contains alleles A and B in

algol13

It is codominant inheritance because, if the placement of the A and B molecules on each cell is controlled by the proteins that are coded by different versions of the same gene, then IA and IB are codominant but both are dominant to Io. If a person receives an IA allele and a IB allele, their blood type is type AB, in which characteristics of both A and B antigens are expressed.

6 0

2 years ago

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Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The

qaws [65]

Answer:

$E_{ionization}=5.45\times 10^{-19}\ J$

Explanation:

$E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

So, $n_i=2$ and $n_f=\infty$ (As the hydrogen has to ionize)

Thus,

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2} - \dfrac{1}{{\infty}^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2})\ J$

$E_{ionization}=5.45\times 10^{-19}\ J$

4 0

3 years ago