The law of conservation of energy has not been broken, provided energy is released from the fission process.
<h3>What is the law of conservation of energy?</h3>
The law states that the total energy of a process is conserved. That is, the total energy or mass of a system before and after undergoing processing remains the same. However, some of the mass/energy can be converted to another form.
When a material undergoes fission, the sum total of the mass of the particles formed should be equal to the mass of the starting materials, provided that all other things remain the same.
However, if energy is released from the fission process, it means that some of the mass of the starting materials has been converted to energy and released to the environment.
More on the law of conservation of energy can be found here: brainly.com/question/20971995
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Answer:
_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)
Explanation:
Step 1:
The unbalanced equation:
AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
Step 2:
Balancing the equation.
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
The above equation can be balanced as follow:
There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)
There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)
Now the equation is balanced
Answer:
The solubility of the mineral compound X in the water sample is 0.0189 g/mL.
Explanation:
Step 1: Given data
The volume of water sample = 46.0 mL.
The weight of the mineral compound X after evaporation, drying, and washing = 0.87 g.
Step 2: Calculate the solubility of X in water
46.00 mL of water sample contains 0.87 g of the mineral compound X.
To calulate how many grams of the mineral compound 1.0 mL of water sample contains:
0.87 g/46.0 mL = 0.0189 g.
This means the solubility of the mineral compound X in the water sample is 0.0189 g/mL.
I think the best answer is the last option. A scatter plot is the appropriate type of graph for the student to use to show the percent samples per group. This plot is somewhat similar to line graphs. However, they are use for a specific purpose which is to show the relationship between two parameters. In this case, the correlation between pH and the percent of samples.
The molecular weight of hemoglobin can be calculated using osmotic pressure
Osmotic pressure is a colligative property and it depends on molarity as
πV = nRT
where
π = osmotic pressure
V = volume = 1mL = 0.001 L
n = moles
R = gas constant = 0.0821 L atm / mol K
T = temperature = 25°C = 25 + 273 K = 298 K
Putting values we will get value of moles
we know that
Therefore
