Answer:
The answer to your question is Molarity = 0.41
Explanation:
Data
mass of KNO₃ = 76.6 g
volume = 1.84 l
density = 1.05 g/ml
Process
1.- Calculate the molecular mass of KNO₃
molecular mass = 39 + 14 + (16 x 3) = 101 g
2.- Calculate the number of moles
101 g of KNO₃ --------------- 1 mol
76.6 g of KNO₃ ------------ x
x = (76.6 x 1) / 101
x = 0.76 moles
3.- Calculate molarity
Molarity = 
Substitution
Molarity = 
Result
Molarity = 0.41
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)
Answer:
47911.1 pa
Explanation:
The SI base unit of pressure is pascal, which is N/m^2.
2200 kg is 2200*9.8=21560 N, and 4500 cm^2=4500/10000=0.45 m^2.
So the total pressure exerted on the ground (!!) is 21560/0.45= 47911.1 Pa.
Answer: <span>The molecules of a substance which must have the
<u>a</u></span>
<u>bility to move past one another</u> are said to be flexible.
Explanation: Those substances are said to be flexible which can be
bent without breaking. There are many substances which are
hard in nature but still can be bent. The hardness of such materials is due to
strong interactions between the molecules and the flexibility comes due to their
amorphous backbone. Therefore, greater the
crystalline level of macromolecules lesser is the flexibility and greater the amorphous character greater is the flexibility and vice versa. Also, the flexibility of polymers is increased by adding
plastisizers in it. Plastisizers make the hard polymers flexible by breaking the crosslinkers and enabling the macromolecules to move past one another.
Answer:
The metal is Tin (Sn)
Explanation:
The following data were obtained from the question:
Mass of metal = 32.56 g
Volume of water = 14.78 mL
Volume of water + metal = 20.44 mL
Next, we shall determine the volume of the metal. This is illustrated below:
Volume of water = 14.78 mL
Volume of water + metal = 20.44 mL
Volume of metal =..?
Volume of metal = (Volume of water + metal) – (Volume of water )
Volume of metal = 20.44 – 14.78
Volume of metal = 5.66mL
Next, we shall determine the density of the metal.
Density of a substance is defined as the mass of the substance per unit volume of the substance. Mathematically, the density of a substance is expressed as:
Density = Mass / volume
With the above formula, we can obtain the density of the metal as follow:
Mass of metal = 32.56 g
Volume of metal = 5.66mL
Density =.?
Density = Mass /volume
Density = 32.56g/5.66mL
Density of the metal = 5.75g/mL
Comparing the density of the metal with standard density of elements, the metal is Tin (Sn)