V1M1 = V2M2
<span>V1 × 2.5 = 1 × 0.75,
so V1 = 0.75/2.5
= 0.3 </span>
Ethane is an alkane. Methane is also an alkane and is considered to be the simplest alkane. The difference is ethane has only 2 carbon. That carbon has 6 hydrogen attached to it. So what we do is we multiply the moles of ethane by the number of hydrogen (by dimension analysis) resulting to 82.68 moles H.
I don't know how well known/accepted this is (it's in my textbook so I'm guessing it's right), but Sulphur has two forms - the alpha and beta forms ,apparently gamma sulphur exists as well.
The alpha form is rhombic, yellow in color and has a MP of 385.8 K. The beta form is colorless and has a MP of 393 K and is formed by melting rhombic sulphur and cooling it till a crust forms on top. Poke a hole and pour out the liquid inside and you get beta sulphur. The transition point is 369K - below it, alpha sulphur is stable and above it, beta sulphur is stable. Both have helped. I had to pull out an old textbook and that's something that I don't usually do.
The amount of the 240 g sample of the radioisotope that will remain after 525 billion years is 7.5 g
<h3>How to the number of half-lives that has elapsed</h3>
- Half-life (t½) = 105 billion years
- Time (t) = 525 billion years
- Number of half-lives (n) = ?
n = t / t½
n = 525 / 105
n = 5
<h3>How to determine the amount remaining</h3>
- Original amount (N₀) = 240 g
- Number of half-lives (n) = 5
- Amount remaining (N) = ?
N = N₀ / 2ⁿ
N = 240 / 2⁵
N = 240 / 32
N = 7.5 g
Learn more about half life:
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start the balancing by writing down how many atoms there are per element. we’ll use this as an example:
C3H8 + O2 --> H2O + CO2
C = 3 C = 1
H = 8 H = 2
O = 2 O = 3
balance the carbon first, as it is easiest to do. add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon atoms on the left of the equation:
C3H8 + O2 --> H2O + (3)CO2
now there are 3 carbon atoms on each side. however, when you do this, you multiply the amount of oxygen atoms you had. therefore, now, there are 6 carbon atoms in 3CO2, plus that other oxygen atom in H2O. you now have 7 O atoms instead of 3.
C = 3 C = 3
H = 8 H = 2
O = 2 O = 7
now let’s move on to the hydrogen atoms.
C3H8 + O2 --> H2O + 3CO2
you have 8 hydrogen atoms on the left side, and 2 on the right. in order to balance them, you have to multiply the right side’s hydrogen atoms by 4. 4(2) = 8.
C3H8 + O2 --> (4)H2O + 3CO2
now both hydrogen and carbon atoms are balanced. same amount on both sides. however, your oxygen atoms have changed due to the multiplying (right side). you now have 10 of them.
C = 3 C = 3
H = 8 H = 8
O = 2 O = 10
now we balance the oxygen atoms. multiply the left side of the equation’s oxygen atoms by 5. 5(2) = 10
C3H8 + (5)O2 --> 4H2O + 3CO2
the chemical equation is all balanced. basically, just multiply with numbers until it equals the same amount on both sides.
C = 3 C = 3
H = 8 H = 8
O = 10 O = 10