The balanced chemical reaction is:
<span>2H2+O2-->2H2O
</span>
To determine the mass of hydrogen that is needed, we need use the initial amount of oxygen and relate it to hydrogen from the reaction given. We do as follows:
192 g O2 ( 1 mol O2 / 32 g O2) ( 2 mol H2 / 1 mol O2 ) ( 2.02 g H2 / 1 mol H2 ) = 24.24 g H2
Which sequence repesent matter that is losing energy
answer: solid liquid gas
HOPE THIS HELPS :)
Explanation:
The end goal of this question is to find density. Density can be found using the following formula:
D=m/v
D=Density
m=mass
v=volume
Therefore we need mass and volume to find density. Mass is already given in the question as 142g. All we have left is volume. When a solid is placed in water in a graduated cylinder the volume goes up. This means if you subtract the volume after the metal was placed from before it was placed, you can find the volume of the metal. Therefore the volume of the metal is 40mL-20mL=20mL. However typically, volume is found in liters so the volume should be .04L-.02L=0.02L. Now that we have volume and mass we can find density using the formula d=m/v. So 142/.02=7,100 grams per liget or 7.1 kilograms per liter. This answer seems unrealistic so I would double check with your teacher about the question.
Hope I helped, 2Trash4U
Answer:
1.208x10⁻³M and 392.5ppm La(NO3)3
Explanation:
The reaction that occurs is:
La2O3 + 6HNO3 → 2La(NO3)3 + 3H2O
Molarity is defined as the moles of solute (In this case, LaO3) per liter of solution. And ppm, are mg of solute per liter of solution.
To solve this question we must find the moles of La(NO3)3 produced and its mass in milligrams to find molarity and ppm:
<em>Moles La2O3 -Molar mass: 325.81g/mol-</em>
0.1968g * (1mol / 325.81g) = 6.04x10⁻⁴ moles La2O3
<em>Moles La(NO3)3:</em>
6.04x10⁻⁴ moles La2O3 * (2mol La(NO3)3 / 1mol La2O3) = 1.208x10⁻³ moles La(NO3)3
<em>Molarity:</em>
1.208x10⁻³ moles La(NO3)3 / 1L =
<h3>1.208x10⁻³M</h3>
<em>Mass La(NO3)3 -Molar mass: 324.92g/mol-</em>
1.208x10⁻³ moles La(NO3)3 * (324.92g / mol) = 0.392.5g La(NO3)3
In mg:
392.5mg La(NO3)3 / 1L =
392.5ppm La(NO3)3