Answer: The pressure of the tires after driving is 4.9atm
Explanation: Please see the attachments below
Option C: I and II only.
The atoms have subparticles and the isotops are atoms of the same element that differ in the number of neutrons.
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)
Hey there !
Molar mass carbon dioxide:
CO2 = 44.01 g/mol
1) number of moles :
1 mole CO2 ------------- 44.01 g
(moles CO2) ------------ 243.6 g
moles CO2 = 243.6 * 1 / 44.01
moles CO2 = 243.6 / 44.01
=> 5.535 moles of CO2
Therefore:
1 mole -------------------- 6.02x10²³ molecules
5.535 moles ------------ ( molecules CO2)
molecules CO2 = 5.535 * ( 6.02x10²³) / 1
=> 3.33x10²⁴ molecules of CO2
Answer:
1. Saturated hydrocarbons may be cyclic or acyclic molecules.
2. An unsaturated hydrocarbon molecule contains at least one double bond.
Explanation:
Hello,
In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).
Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true
Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.
Therefore, first and second statements are correct.
Best regards.
