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rewona [7]
3 years ago
11

In the article, Invasion of the Periodical Cicadas, why do scienctists believe that cicadas only reproduce every 13 or 17 years?

Explain your answer with evidence from the text.
Chemistry
1 answer:
frez [133]3 years ago
8 0

Hi, you've asked an incomplete question. However, I assumed you are referring to the article found on the Scientific American website.

Explanation:

<em>Remember,</em> according to that article we are told that scientists notice that these insects have a long nymphal (immature form before becoming adults) stage, one that can last up to 13 to 17 years on the ground before they leave the ground looking for mating partners.

Because it is only after mating occurs at this point that their eggs are laid, that is why scientists believe that cicadas only reproduce every 13 or 17 years.

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Which of the following is the correct molecule for dinitrogen trisulfide?
Setler79 [48]

Answer:

A. N2S3

yep u r right

Explanation:

Dinitrogen Trisulfide N2S3 Molecular Weight -- EndMemo.

4 0
3 years ago
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Which of the following is true about two neutral atoms of the element gold?
Nuetrik [128]
C is correct on this problem
7 0
3 years ago
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GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide g
JulsSmile [24]

Answer:

The equation to show the the correct form to show the standard molar enthalpy of formation:

\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of H_2 gas and 1 mole of Br_2 liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

H_2(g) + Br_2(l)\rightarrow 2HBr(g) ,\Delta H_{f}^o= -72.58kJ

Divide the equation by 2.

\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ

The equation to show the the correct form to show the standard molar enthalpy of formation:

\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ

4 0
3 years ago
At a given temperature, 2.6 atm of h2 and 3.14 atm of cl2 are mixed and allowed to come to equilibrium. the equilibrium pressure
Keith_Richards [23]

The solution would be like this for this specific problem:

<span>Given:

H2 = </span><span>2.6 atm
CL2 = 3.14 atm</span>

 

<span>
pressure H2 = 2.6 - x 
pressure Cl2 = 3.14 - x 
<span>pressure HBr = 2x = 1.13

x = 1.13 / 2 = 0.565 

<span>pressure H2 = 2.6 - 0.565 = 2.035
pressure Br2 = 3.14 - 0.565 = 2.575 

Kp = (1.13)^2 / 2.035 x 2.575</span></span></span>

 

= 1.2769 / (5.240125)

= 0.24367739319195629875241525726963

= 0.244

<span>Therefore, the Kp for the reaction at the given temperature is 0.244.

To add, </span>the hypothetical pressure of a gas if it alone occupied the whole volume of the original mixture at the same temperature is called the partial pressure or Kp.

7 0
3 years ago
If the percent yield for the following reaction is 75.0%, and 25.0 g of NO₂ are consumed in the reaction, how many grams of nitr
victus00 [196]

Answer:

17.1195 grams of nitric acid are produced.

Explanation:

3NO_2+H_2O\rightarrow 2HNO_3+NO

Moles of nitrogen dioxide :

\frac{25.0 g}{56 g/mol}=0.5434 mol

According to reaction 3 moles of nitrogen dioxides gives 2 moles of nitric acid.

Then 0.5434 moles of nitrogen dioxides will give:

\frac{2}{3}\times 0.5434 mol=0.3623 mol of nitric acid.

Mass of 0.3623 moles of nitric acid :

0.3623 mol\times 63 g/mol=22.8260 g

Theoretical yield = 22.8260 g

Experimental yield = ?

\%Yield=\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

75\%=\frac{\text{Experimental yield}}{22.8260 g}

Experimental yield of nitric acid = 17.1195 g

7 0
3 years ago
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