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VMariaS [17]
3 years ago
6

Match each processes with the correct transfer of energy.

Chemistry
1 answer:
sveticcg [70]3 years ago
7 0

freezing-energy lost (exothermic)

sublimation-energy gain (endothermic)

evaporation- energy gain(endothermic)

Melting- energy gain(endothermic)

deposition- energy lost(exothermic)

condensation-energy lost(exothermic)

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15. If you dilute a 6 M solution of HCl from 5 mL to 50mL, what is the concentration of this new solution? (M1V1 = M2V2)
Andreyy89

Answer:

B) 0.6M

Explanation:

               I apologize in advance if it is not correct :l

The (M1V1= M2V2) is given for you to plug in the correct numbers so let's jot this down.

                    (M1*V1= M2*V2)

so they give us 6M which would be our (M1), from this we can also conclude that 5mL is also V1; ( if you notice the M1's and V1's are always found next to eachother). This leads us to our 50mL, this would be our V2 because the volume went from 5mL to 50mL. Now lets put this in order based on what we know.

M1= 6M                                (M1*V1= M2*V2)

V1= 5mL

M2= ?

V2= 50mL

now we plug in what we know into the equation to find the unknown (M2)

                   (6M*5mL= M2*50mL)

now we could do the long math, but I don't think your on brainly to do the hard way. so lets keep it simple!

                   We are going to put the 50mL under the (6M*5mL) for division.

                             \frac{(6M*5mL)}{(50mL)} This is honestly MUCH easier, than manually answering. you just put that in the calculator and it'll give you B) 0.6M

                 

   honestly though I might not know what I'm doing cuz im currently doing my test and decided to answer this question ;)

        Good Luck!

6 0
3 years ago
A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What i
In-s [12.5K]

Answer:

81.26% is the percent yield

Explanation:

Based on the reaction:

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

<em>Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.</em>

<em />

To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:

Actual yield (0.366g) / Theoretical yield * 100

<em>Moles CaCl₂ = Moles CaCO₃:</em>

0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃

<em>Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:</em>

0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃

Percent yield = 0.366g / 0.450g * 100

81.26% is the percent yield

3 0
3 years ago
The article is on newsela called the water cycle
garri49 [273]

Answer:

Explanation:Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?

7 0
3 years ago
What is 57.048 rounded to three significant figures?
AnnyKZ [126]
57.0 is it rounded to three sig figs. You count three spaces then round from there, which would be the zero and you would round down because the four is there.
4 0
3 years ago
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
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