Answer:
B 1.23 g/cc
Explanation:
For something to float on seawater, the density must be less than 1.03 g/mL. If the object sinks, the density is greater than 1.03 g/mL.
Let’s examine the answer choices. Keep in mind, the ice berg is mostly below the water level.
A. 0.88 g/cc
This is less than 1.03 g/cc, which would result in floating.
B. 1.23 g/cc
This is the best answer choice. The iceberg is mostly beneath the water, but some of it is exposed. The density is greater than 1.03 g/mL, but not so much greater that it would immediately sink.
C. 0.23 g/cc
This is less than 1.03 g/cc, which would produce floating.
D. 4.14 g/cc
This is much greater than 1.03 g/cc and the result would be sinking.
1) we calculate the molar mass of He (helium) and Kr (Krypton).
atomic mass (He)=4 u
atomic mas (Kr)=83.8 u
Therefore the molar mass will be:
molar mass(He)=4 g/mol
molar mass(Kr)=83.8 g/mol.
1) We can find the next equation:
mass=molar mass x number of moles.
x=number of moles of helium
y=number of moles of helium.
(4 g/mol) x +(83.8 g/mol)y=103.75 g
Therefore, we have the next equation:
(1)
4x+83.8y=103.75
2) We can find other equation:
We have 30% helium atoms and 70% Kryptum atoms, therefore we have 30% Helium moles and 70% of Krypton moles.
1 mol is always 6.022 * 10²³ atoms or molecules, (in this case atoms).
Then:
x=number of moles of helium
y=number of moles of helium.
(x+y)=number of moles of our sample.
x=30% of (x+y)
Therefore, we have this other equation:
(2)
x=0.3(x+y)
With the equations(1) and (2), we have the next system of equations:
4x+83.8y=103.75
x=0.3(x+y) ⇒ x=0.3x+0.3y ⇒ x-0.3x=0.3y ⇒ 0.7 x=0.3y ⇒ x=0.3y/0.7
⇒x=3y/7
We solve this system of equations by substitution method.
x=3y/7
4(3y/7)+83.8y=103.75
lower common multiple)7
12y+586.6y=726.25
598.6y=726.25
y=1.21
x=3y/7=3(1.21)/7=0.52
We have 0.52 moles of helium and 1.21 moles of Krypton.
1 mol=6.022 * 10²³ atoms
Total number of particles=(6.022 *10²³ atoms /1 mol) (number of moles of He+ number of moles of Kr).
Total number of particles=6.022 * 10²³ (0.52+1.21)=6.022 * 10²³ (1.73)=
=1.044 * 10²⁴ atoms.
Answer: The sample have 1.044 * 10²⁴ atoms.
Physical changes occur when the properties of a substance are retained and/or the materials can be recovered after the change. Chemical changes involve the formation of a new substance. Formation of a gas, solid, light, or heat are possible evidence of chemical change.
Answer:
The final volume is 39.5 L = 0.0395 m³
Explanation:
Step 1: Data given
Initial temperature = 200 °C = 473 K
Volume = 0.0250 m³ = 25 L
Pressure = 1.50 *10^6 Pa
The pressure reduce to 0.950 *10^6 Pa
The temperature stays constant at 200 °C
Step 2: Calculate the volume
P1*V1 = P2*V2
⇒with P1 = the initial pressure = 1.50 * 10^6 Pa
⇒with V1 = the initial volume = 25 L
⇒with P2 = the final pressure = 0.950 * 10^6 Pa
⇒with V2 = the final volume = TO BE DETERMINED
1.50 *10^6 Pa * 25 L = 0.950 *10^6 Pa * V2
V2 = (1.50*10^6 Pa * 25 L) / 0.950 *10^6 Pa)
V2 = 39.5 L = 0.0395 m³
The final volume is 39.5 L = 0.0395 m³
Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)
