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natka813 [3]
3 years ago
6

The energy of the universe is constant. That what mentioned by the first law of thermodynamics,

Chemistry
1 answer:
Goryan [66]3 years ago
5 0

Answer:

No, ΔE does not always equal zero because it refers to the systems internal energy, which is affected by heat and work

Explanation:

According to the first law of thermodynamics, energy is neither created nor destroyed. This implies that the total energy of a system is always a constant.

So, according to the first law of thermodynamics we have that ΔE = q + w. This means that the value of  ΔE depends on q (heat) and  w(work). Hence ΔE is not always zero since it depends on the respective values of q and w.

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A hydrocarbon with vapour density
Dafna1 [17]

Answer:

the answer is C4H10 I hope I helped

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3 years ago
What is the single factor that classifies an element as a transition metal? Question 16 options:
alukav5142 [94]
A...Just got it right on my exam:)
4 0
3 years ago
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Everything in group 2 on the periodic table are called
Ludmilka [50]

Answer:

group two elements are called alkaline earth metals

Explanation:

Because their oxides form in the earth and are water soluble

3 0
3 years ago
Radioactive radium has a half-life of approximately 1,599 years. The initial quantity is 13 grams. How much (in grams) remains a
Luda [366]

The quantity of substance remains after 850 years is 8.98g if the half life of radioactive radium is 1,599 years.

<h3>What is half life period? </h3>

The time taken by substance to reduce to its half of its initial concentration is called half life period.

We will use the half- life equation N(t)

N e^{(-0.693t) /t½}

Where,

N is the initial sample

t½ is the half life time period of the substance

t2 is the time in years.

N(t) is the reminder quantity after t years .

Given

N = 13g

t = 350 years

t½ = 1599 years

By substituting all the value, we get

N(t) = 13e^(0.693 × 50) / (1599)

= 13e^(- 0.368386)

= 13 × 0.691

= 8.98

Thus, we calculated that the quantity of substance remains after 850 years is 8.98g if the half life of radioactive radium is 1,599 years.

learn more about half life period:

brainly.com/question/20309144

#SPJ4

4 0
1 year ago
How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)
Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide =\frac{100 g}{44 g/mol}=2.273 moles

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = 2.273\times 10^{-3} kmol

2) 1 liter of ethyl alcohol of density 0.789 g/cm^3

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = 0.789 g/cm^3

1 cm^3=1 mL

Mass of ethyl alcohol = m

m=d\times V=0.789 g/cm^3\times 1000 mL=789 g

Molar mass of  ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = \frac{789 g}{46 g/mol}=17.152 mol

17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol

3) Volume of oxygen gas,V =1.5 m^3=1500 L

1 m^3= 1000 L

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

PV=nRT

n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol

0.01632 mol = 0.01632 × 0.001 kmol=1.632\times 10^{-5} kmol

4) Volume water in mixture = 1 L

Density of water =  1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L

Mass of water = 1000 g/L\times 1 L = 1000 g

Volume of alcohol = 2.5 L

Density of alcohol =  789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L

Mass of alcohol = 789 g/L\times 2.5 L = 1972.5 g

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

\frac{1000 g}{2972.5 g}\times 100=33.64\%

Mass percentage of alcohol :

\frac{1972.5 g}{2972.5 g}\times 100=66.36\%

Moles of water :

n_1=\frac{1000 g}{18 g/mol}=55.55 mol

Moles of alcohol =

n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol

Mole fraction of water :

\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644

Mole fraction of alcohol :

\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356

3 0
3 years ago
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