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3 years ago

10

Select the correct answer. A light source emits white light. A red-colored filter is fitted over the light. What color will a pe

rson looking at the filtered light see?
white light
red light
pink light
no light

2 answers:

liq [111]3 years ago

7 0

Im pretty sure it’s pink light :)

Rainbow [258]3 years ago

7 0

Answer:

It's a red light

Explanation:

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Which of the following is true?

Tomtit [17]

Answer:

a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.

Explanation:

The change in free energy (ΔG) that is, the <u>energy available to do work</u>, of a system for a constant-temperature process is:

$ΔG = ΔH - TΔS$

When ΔG < 0 the reaction is spontaneous in the forward direction.

When ΔG > 0 the reaction is nonspontaneous. The reaction is

spontaneous in the opposite direction.

When ΔG = 0 the system is at equilibrium.

If <u>both ΔH and ΔS are positive</u>, then ΔG will be negative only when the TΔS term is greater in magnitude than ΔH. This condition is met when T is large.

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3 years ago

Which is the correct equation for the reaction of magnesium with hydrochloric acid to produce hydrogen gas and magnesium chlorid

djyliett [7]

Answer:

Mg(s) + 2 HCl(aq) --> MgCl 2(aq) + H 2(g)

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2 years ago

Calculate the enthalpy of combustion of 1 mol decane, C10H22, (l), to form CO2 and H2O. ∆Hf0 for decane is —300.9 kJ/mol using t

worty [1.4K]

Answer:

heyy do you got the answer? i got that cuestion but i need the answer!!! did you finally get it?

Explanation:

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3 years ago

An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder

seraphim [82]

Answer:

$\Delta H=-11897J$

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

$\Delta H=\Delta U+V\Delta P$

Whereas the change in the internal energy is computed by:

$\Delta U=nCv\Delta T$

So we compute the initial and final temperatures for one mole of the ideal gas:

$T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}$

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

$\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J$

Then, the volume-pressure product in Joules:

$V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J$

Finally, the change in the enthalpy for the process:

$\Delta H=-7140J-4757J\\\\\Delta H=-11897J$

Best regards.

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3 years ago

PLS HELP ASAP WILL MAKE BRAINLIST

Delvig [45]

Answer:

2y+3m2= 2ym3

Explanation:

Balances out

2y=2y

6m=6m

6 0

2 years ago