Answer:
p-Nitro acetanilide. 2-Naphthol aniline dye.
The new pressure would be = 4.46 atm
<h3>Further explanation</h3>
Given
V₁=6.7 L(at STP, 1 atm 273 K)
V₂=1.5 L
Required
The new pressure
Solution
Boyle's Law
At a constant temperature, the gas volume is inversely proportional to the pressure applied
![\rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}](https://tex.z-dn.net/?f=%5Crm%20p_1V_1%3Dp_2.V_2%5C%5C%5C%5C%5Cdfrac%7Bp_1%7D%7Bp_2%7D%3D%5Cdfrac%7BV_2%7D%7BV_1%7D)
P₂ = (P₁V₁)/V₂
P₂ = (1 atm x 6.7 L)/1.5 L
P₂ = 4.46 atm
The SI base units and their physical quantities are the metre for measurement of length, the kilogram for mass, the second for time, the ampere for electric current, the kelvin for temperature, the candela for luminous intensity, and the mole for amount of substance.
Answer:
2 C2H2 + 5 O ---> 4 CO2 + 2 H2O
Explanation:
combustion reactions always end with CO2 + H2O
and you can use this website to balance out equations when you're stuck
https://en.intl.chemicalaid.com/tools/equationbalancer.php?equation=C2H2+%2B+O2+%3D+CO2+%2B+H2O
The answer is 3.
<span>The relation between number of half-lives (n) and decimal amount remaining (x) can be expressed as:
</span>
![(1/2) ^{n} =x](https://tex.z-dn.net/?f=%281%2F2%29%20%5E%7Bn%7D%20%3Dx)
We need to calculate n, but we need x to do that. To calculate what p<span>ercentage of a radioactive species would be found as daughter material, we must calculate what amount remained:
1.28 -</span> 1.12 = 0.16
If 1.28 is 100%, how much percent is 0.16:
1.28 : 100% = 0.16 : x
x = 12.5%
Presented as decimal amount:
x = 0.125
Now, let's implement this in the equation:
<span>
![(1/2) ^{n} =0.125](https://tex.z-dn.net/?f=%281%2F2%29%20%5E%7Bn%7D%20%3D0.125)
</span>
Because of the exponent, we will log both sides of the equation:
![n * log(1/2) = log(0.125)](https://tex.z-dn.net/?f=n%20%2A%20log%281%2F2%29%20%3D%20log%280.125%29)
![n = \frac{log(0.125)}{log(1/2)}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7Blog%280.125%29%7D%7Blog%281%2F2%29%7D%20)
<span>
![n = \frac{log(0.125)}{log(0.5)}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7Blog%280.125%29%7D%7Blog%280.5%29%7D%20)
</span>
![n= \frac{-0.903}{-0.301}](https://tex.z-dn.net/?f=n%3D%20%5Cfrac%7B-0.903%7D%7B-0.301%7D%20)
![n = 3](https://tex.z-dn.net/?f=n%20%3D%203)
Therefore, 3 half-lives have passed <span> since the sample originally formed.</span>