Answer:
d
Explanation:
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Explanation:
It's (D), nuclear fission................
Answer:
8.8g of Al are necessaries
Explanation:
Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.
To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:
<em>Moles H2:</em>
PV = nRT; PV/RT = n
<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>
Replacing:
1atm*11L/0.082atmL/molK*273.15K = n
n = 0.491 moles of H2 must be produced
<em />
<em>Moles Al:</em>
0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required
<em />
<em>Mass Al -Molar mass: 26.98g/mol-:</em>
0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries
Answer:
E = 147000 J
Explanation:
Given that,
The mass of meteor, m = 50 kg
The altitude of the meteor, h = 300 m
We need to find the potential energy of the meteor. The formula for the potential energy is given by :
Put all the values,
So, the required potential energy is equal to 147000 J.
Supposing a temperature of 25 degrees and supposing that all
activity coefficients are 1
pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14
Thus a pH of 2.50 would mean that the [H+], the concentration of the hydrogen
ion, would be 10^(-2.50)
pH + pOH = 14
pOH = 14 - pH = 14 - 2.5 = 11.5
MOH- levels would be coordinated with pOH
pOH = -log[OH-] ==> [OH-] = [MOH-] = 10^-pOH = 10^-11.5 = 3.2 x 10^-12
Therefore, MOH¯ = 3.2 × 10¯12 M
