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3 years ago

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When boiling a pot of water on the stove without a lid on the pot, the _____________. A energy and water molecules are free to e

scape into the surrounding environment. B molecules can escape out into the environment, but the energy are forced to stay in the system. C energy can escape out into the environment, but the molecules are forced to stay in the system. D both the energy and molecules are trapped in the system.

2 answers:

Allisa [31]3 years ago

4 0

Answer: Its A

Explanation: I got an A+

ch4aika [34]3 years ago

3 0

Answer:

A) Energy and Water molecules are free to escape into the surrounding environment.

Explanation:

<em>Hello! This is the correct answer! (I took the K12 test) </em>

<em>I hope this helps! Have a blessed day! :)</em>

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Unit: Chemical Quantities

Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

$n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 1.993\,moles$

2) The number of atoms of helium is:

$x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)$

$x = 9.033\times 10^{23}\,atoms$

3) The quantity of moles of carbon monoxide is:

$n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.689\,moles$

4) The number of molecules of sulfur dioxide is:

$x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.355\times 10^{24}\,molecules$

5) The quantity of moles of sodium chloride is:

$n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.399\,moles$

6) The number of formula units of magnesium iodide is:

$x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)$

$x = 1.084\times 10^{24}\,f.u.$

7) The quantity of moles of potassium permanganate is:

$n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }$

$n = 1.214\,moles$

8) The number of molecules of carbon tetrachloride is:

$x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.506\times 10^{23}\,molecules$

9) The quantity of moles of aluminium is:

$n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 0.609\,moles$

10) The number of molecules of oxygen difluoride is:

$x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 2.120\times 10^{24}\,molecules$

3 0

3 years ago

A student makes the following hypothesis: If the amount of chocolate increases then the cake will taste sweeter. What is the ind

Talja [164]

Answer:

the chocolate

Explanation:

You are changing the amount of chocolate

8 0

3 years ago

Hellllppppppp meeee fassttttt please

Volgvan

a tick attaches itself to the skin of a dog

5 0

3 years ago

Atrial natriuretic peptide is a hormone that controls blood pressure in part by increasing the urinary excretion of sodium. True

mariarad [96]

Answer:

The answer is "true"

Explanation:

ANP, or ANF, is a human peptide hormone, that controls your blood pressure, a cardiac atrium secretion hormone. It response primarily to the volume and stress expansion of the auricles and ventricles.

ANP's main function decreases the amount of increased extracellular fluid by increasing renal excretion.
This hormone is primarily developed by cardiovascular myocytes and, emanates from the wall tension locally, that's why the given statement is true.

4 0

3 years ago

Match each substance with the correct designation for the equation HSO3- + CH3NH2 <=> SO32- + CH3NH3+ HSO3- CH3NH2 SO32- C

Zanzabum

Answer:

$HSO_3^-$ : conjugate acid of $SO_3^{2-}$

$CH_3NH_2$ : conjugate base of $CH_3NH_3^+$

$SO_3^{2-}$ : conjugate base of $HSO_3^-$

$CH_3NH_3^+$ : conjugate acid of $CH_3NH_2$

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

$HSO_3^-+CH_3NH_2\rightleftharpoons SO_3^{2-}+CH_3NH_3^+$

Here in forward reaction $CH_3NH_2$ is accepting a proton, thus it is considered as a base and after accepting a proton, it forms $CH_3NH_3^+$ which is a conjugate acid.

And $HSO_3^-$ is losing a proton, thus it is considered as an acid and after loosing a proton, it forms $SO_3^{2-}$ which is a conjugate base.

Similarly in the backward reaction, $CH_3NH_3^+$ is loosing a proton, thus it is considered as a acid and after loosing a proton, it forms $CH_3NH_2$ which is a conjugate base.

And $SO_3^{2-}$ is accepting a proton, thus it is considered as a base and after accepting a proton, it forms $HSO_3^{-}$ which is a conjugate acid.

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3 years ago