Answer:
=> 2.8554 g/mL
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 16.59 g
Volume (v) = 5.81 mL
From our question, we are to determine the density (rho) of the rock.
The formula:
Substitute the values into the formula:
= 2.8554 g/mL
Therefore, the density (rho) of the rock is 2.8554 g/mL.
B. the tilt of the Earth on it's axis
The moon affects the tides by it's gravitational pull on Earth. So the moon has no affect on the earth's seasons. The distance from earth from the sun only affects the ability of Earth to sustain liquid water because earth is in something scientists like to call the "Goldilocks Zone" This has no affect on the seasons. However, the angle at which the sun's light hit the earth does effect seasonal changes and it's tilt does change.
2.1648 kg of CH4 will generate 119341 KJ of energy.
Explanation:
Write down the values given in the question
CH4(g) +2 O2 → CO2(g) +2 H20 (g)
ΔH1 = - 802 kJ
2 H2O(g)→2 H2O(I)
ΔH2= -88 kJ
The overall chemical reaction is
CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ
CH4 +2 O2 → CO2 +2 H20
(1mol)+(2mol)→(1mol+2mol)
Methane (CH4) = 16 gm/mol
oxygen (O2) =32 gm/mol
Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O
which generate 882 KJ /mol
Therefore to produce 119341 KJ of energy
119341/882 = 135.3 mol
to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require
=135.3 *16
=2164.8 gm
=2.1648 kg of CH4
2.1648 kg of CH4 will generate 119341 KJ of energy
When placed in a container, the heaviest (most dense) will sink to the bottom and the lightest (least dense) will rise to the top.
Therefore, Gasoline would rise to the top.
The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of stomach acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 1.25 mL
