The uppermost layer of the atmosphere

What is the main function of the part labeled Y in the model?

Tanya [424]

Answer

D

Explanation

the answer is D

8 0

3 years ago

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A student dissolves a mass of ammonium nitrate in water and notices that the temperature of the mixture decreases by

Mrac [35]

The conclusion of the student would be that the dissolution of ammonium nitrate is endothermic.

<h3>What are endothermic reactions?</h3>

They are reactions in which heat is absorbed from the surrounding. The consequence is that the reaction mixture's temperature is always lower.

Thus, the energy of the products is always higher than the energy of the reactants in endothermic reactions.

For endothermic reactions, the enthalpy change is always positive.

More on endothermic reactions can be found here: brainly.com/question/23184814

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6 0

2 years ago

WILl GIVE BRANILIEST AND 10+ POINTS

Ne4ueva [31]

Answer:

the answer is C. The money time and resources used on the model

5 0

3 years ago

Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical for

elena55 [62]

Answer:

(a) $C_5H_8O_2$

(b) $CHCl$

(c) $CH_2$

(d) $CH$

(e) $C_3H_3N$

Explanation:

Hello,

(a) For the lucite, one computes the moles of C, H and O that are present:

$n_C=0.599gC*\frac{1molC}{12gC}=0.05molC\\n_H=0.0806gH*\frac{1molH}{1gH}=0.0806molH\\n_O=0.32gO*\frac{1molO}{16gO}=0.02molO\\$

Now, dividing each moles by the smallest moles (oxygen's moles), one obtains:

$C=\frac{0.05}{0.02} =2.5;H=\frac{0.0806}{0.02} =4;O=\frac{0.02}{0.02} =1$

Finally, we look for the smallest whole number subscript by multiplying by 2, so the empirical formula turns out into:

$C_5H_8O_2$

(b) For the Saran, one computes the moles of C, H and Cl that are present:

$n_C=0.248gC*\frac{1molC}{12gC}=0.021molC\\n_H=0.02gH*\frac{1molH}{1gH}=0.02molH\\n_{Cl}=0.731gCl*\frac{1molCl}{35.45gCl}=0.021molCl\\$

Now, dividing each moles by the smallest moles (hydrogen's moles), one obtains:

$C=\frac{0.021}{0.02} =1;H=\frac{0.02}{0.02} =1;Cl=\frac{0.021}{0.02} =1$

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

$CHCl$

(c) For the polyethylene, one computes the moles of C and H that are present:

$n_C=0.86*\frac{1molC}{12gC}=0.072molC\\n_H=0.14gH*\frac{1molH}{1gH}=0.14molH$

Now, dividing each moles by the smallest moles (carbon's moles), one obtains:

$C=\frac{0.072}{0.072} =1;H=\frac{0.14}{0.072} =2$

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

$CH_2$

(d) For the polystyrene, one computes the moles of C and H that are present:

$n_C=0.923*\frac{1molC}{12gC}=0.077molC\\n_H=0.077gH*\frac{1molH}{1gH}=0.077molH$

Now, dividing each moles by the smallest moles (either carbon's or hydrogen's moles), one obtains:

$C=\frac{0.077}{0.077} =1;H=\frac{0.077}{0.077} =1$

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

$CH$

(e) For the orlon, one computes the moles of C, H and N that are present:

$n_C=0.679*\frac{1molC}{12gC}=0.057molC\\n_H=0.057gH*\frac{1molH}{1gH}=0.057molH\\n_N=0.264gN*\frac{1molN}{14gN}=0.019molN$

Now, dividing each moles by the smallest moles (nitrogen's moles), one obtains:

$C=\frac{0.057}{0.019} =3;H=\frac{0.057}{0.019} =3;N=\frac{0.019}{0.019} =1$

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

$C_3H_3N$

Best regards.

3 0

4 years ago

An open flask sitting in a lab refrigerator looks empty, but it is actually filled with a mixture of gases called air. If the fl

erastova [34]

Answer: $67.4\times 10^{23}$ molecules are there in the flask

Explanation:

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 1 atm

V= Volume of the gas = 2.00 L

T= Temperature of the gas = 273 K $0^00C=273K$

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas

$n=\frac{RT}{PV}=\frac{0.0821\times 273}{1\times 2.00}=11.2$

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of gaseous air contains = $6.023\times 10^{23}$ molecules

11.2 moles of gaseous air contains = $\frac{6.023\times 10^{23}}{1}\times 11.2=67.4\times 10^{23}$ molecules

$67.4\times 10^{23}$ gaseous molecules are contained in the flask.

7 0

3 years ago