Answer:
The correct statements are:
The rate of disappearance of B is twice the rate of appearance of C.
Explanation:
Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.
3A + 2B → C + 2D
Rate of the reaction:
![R=-\frac{1}{3}\times \frac{d[A]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{1}\times \frac{d[C]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D)
![-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{2}\times \frac{d[D]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
The rate of disappearance of B is twice the rate of appearance of C.
![\frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![2\times \frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{1}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=2%5Ctimes%20%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
<span>The answer is "D" where the number of collisions per unit area is reduced by one-half. Drawing back on the piston means the volume is increased. The pressure is reduced. There are fewer collisions when the pressure is reduced.</span>
<span>0.925 grams if using hydrochloric acid in the reaction.
0.462 grams if using sulfuric acid in the reaction.
0.000 grams if using nitric acid in the reaction.
Assuming you're using HCl or a similar acid for this reaction, the equation for the reaction is:
Zn + 2 HCl ==> ZnCl2 + H2
So each mole of zinc used, produces 1 mole of hydrogen gas, or 2 moles of hydrogen atoms. So we need to look up the atomic weights of both zinc and hydrogen.
Atomic weight zinc = 65.38
Atomic weight hydrogen = 1.00794
Moles zinc = 30.0 g / 65.38 g/mol = 0.458855919 mol
Since we produce 2 moles of hydrogen atoms per mole of zinc, multiply by 2 and the atomic weight of hydrogen to get the mass of hydrogen produced. So
0.458855919 * 2 * 1.00794 = 0.92499847 grams.
Rounding to 3 significant figures gives 0.925 grams.
To show the assumption of the acid used, the balanced equation for sulfuric acid would be
Zn2 + H2SO4 ==> Zn(SO4)2 + H2
Which means that for every mole of zinc used, 1 mole of hydrogen gas is generated (half that produced via hydrochloric acid).
If nitric acid were used, the reaction is
4Zn + 10HNO3 ==> 4Zn(NO3)2 + N2O + 5H2O
Which means that NO hydrogen gas is generated.
The only justification for assuming hydrochloric acid is used is that it's a fairly common acid that's easy to obtain. But as shown above with 2 alternative acids, the amount of hydrogen gas generated is very dependent upon the exact chemical reaction occurring and asking "How many grams of hydrogen are produced if 30.0 g of zinc reacts?" is a rather silly question unless you specify EXACTLY what the reaction is.</span>
Answer:
gold and copper
Explanation:
but I think there is 1 more
Answer:
THE NEW PRESSURE OF THE HELIUM GAS AT 2.98 L VOLUME IS 124.8 kPa.
AT AN INCREASE ALTITUDE, THERE IS A LOWER PRESSURE ENVIRONMENT AND THE HELIUM GAS PRESSURE DECREASES AND HENCE AN INCREASE IN VOLUME.
Explanation:
The question above follows Boyle's law of the gas law as the temperature is kept constant.
Boyle's law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.
Mathematically, P1 V1 = P2 V2
P1 = 150 kPa = 150 *10^3 Pa
V1 = 2.48 L
V2 = 2.98 L
P2 = ?
Rearranging the equation, we obtain;
P2 = P1 V1 / V2
P2 = 150 kPa * 2.48 / 2.98
P2 = 372 *10 ^3 / 2.98
P2 = 124.8 kPa.
The new pressure of the gas when at a height which increases the volume of the helium gas to 2.98 L is 124.8 kPa.
