The engine ran for 10 second
You answer this by using the pH formula and and the relation of pH and pOH, pH = -log[H+] and 14 = pH + pOH. The correct classification are as follows:
<span>A. [H2O+]=6.0x10^-12
basic
B. [H3O+]=1.4x10^-9
basic
C. [OH-]=5.0x10^-12
acidic
D. {OH-]=3.5x10^-10
acidic
Hope this answers the question.
</span>
there's no picture here but I guess the answer would be:
considering the constant temperature, if you double the volume, the pressure would be halved.
like: volume is 2, pressure is 4
if 2×2, then:4÷2
Answer:
m = 4.7 μg
Explanation:
Given data:
density of acetone = 60.0 μg/L
Volume = 79.0 mL
Mass = ?
Solution:
Formula:
d = m/v
v = 79.0 mL × 1L /1000 mL
v = 0.079 L
Now we will put the values on formula:
d = m/v
60.0 μg/L = m/0.079 L
m = 60.0 μg/L × 0.079 L
m = 4.7 μg
So health risk limit for acetone = 4.7 μg
CxHy + O2 --> x CO2 + y/2 H2O
Find the moles of CO2 : 18.9g / 44 g/mol = .430 mol CO2 = .430 mol of C in compound
Find the moles of H2O: 5.79g / 18 g/mol = .322 mol H2O = .166 mol of H in compound
Find the mass of C and H in the compound:
.430mol x 12 = 5.16 g C
.166mol x 1g = .166g H
When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.
Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).
In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).
