Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:
In terms of mass, specific heat and temperature change is:
Now, solve for the final temperature, as follows:
Then, plug in the masses, specific heat and temperatures to obtain:
Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.
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Answer:
I think it will carbon hope it helps
Answer:
1.99grams
Explanation:
- First, we need to calculate the molar mass of the compound: Ca(HCO3)2
Ca = 40g/mol, H = 1g/mol, C = 12g/mol, O = 16g/mol
Hence, Ca(HCO3)2
= 40 + {1 + 12 + 16(3)}2
= 40 + {13 + 48}2
= 40 + {61}2
= 40 + 122
= 162g/mol
Molar mass of Ca(HCO3)2 = 162g/mol
- Next, we calculate the mass of oxygen in one mole of the compound, Ca(HCO3)2.
Oxygen = {16(3)}2
= 48 × 2
= 96g of Oxygen
- Next, we calculate the percentage composition of oxygen by mass by dividing the mass of oxygen in the compound by the molar mass of the compound i.e.
% composition of O = 96/162 × 100
= 0.5926 × 100
= 59.26%.
- The number of moles of the compound, Ca(HCO3)2, must be converted to mass by using the formula; mole = mass/molar mass
0.0207 = mass/162
Mass = 162 × 0.0207
Mass = 3.353grams
However, in every gram of Ca(HCO3)2, there is 0.5926 g of oxygen
Hence, in 3.353grams of Ca(HCO3)2, there will be;
0.5926 × 3.353
= 1.986
= 1.99grams.
Therefore, there is 1.99grams of Oxygen in 0.0207 moles (3.353g) of Ca(HCO3)2.
Answer:
sodium fluoride
Explanation:
When we add sodium fluoride to a solution of a given mixture, we get barium fluoride from Ba, which would be an insoluble salt, and silver fluoride from Ag, which would be a soluble salt.
The solubility rule will be used to determine the barium salt that forms as a precipitate and leaves Ag+ salt in the solution.
From the following equations, we will see that the precipitate is formed in Ba but Ag remains dissolved in the solution.
