The six metalloids are boron, silicon, germanium, arsenic, antimony, and tellerium.
Answer:
Carbonic acid could be formed.
Explanation:
Hello,
Based on her claim, it would be a really useful strategy to prevent global warming, nevertheless, there would be a problem if a increasing amount of carbon dioxide is not buried at the bottom of the ocean yet it flows freely along the sea and probably reacting with the water, causing carbonic acid to be formed and subsequently cutting back the sea's pH (increasing its acidity).
It would be useful, but a constant monitoring of the sea's pH must be needed because this could cause some species to be affected not only by the temperature but for the acid pH as well.
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Answer:
K₂CO₃
Explanation:
Given parameters:
Number of moles of K = 0.104mol
Number of moles of C = 0.052mol
Number of moles of O = 0.156mol
Method
From the given parameters, to calculate the empirical formula of the elements K, C and O, we reduce the given moles to the simplest fraction.
Empirical formula is the simplest formula of a compound and it differs from the molecular formula which is the actual formula of a compound.
- Divide the given moles through by the smallest which is C, 0.052mol.
- Then approximate values obtained to the nearest whole number of multiply by a factor to give a whole number ratio.
- This is the empirical formula
Solution
Elements K C O
Number of moles 0.104 0.052 0.156
Dividing by the
smallest 0.104/0.052 0.052/0.052 0.156/0.052
2 1 3
The empirical formula is K₂CO₃
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Answer:</h3>
200 mL
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Explanation:</h3>
Concept tested: Dilution formula
We are given;
- Concentration of stock solution as 1.00 M
- Volume of the stock solution as 50 mL
- Molarity of the dilute solution as 0.25 M
We are required to calculate the volume of diluted solution;
- The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
- Using the dilution formula we can determine the volume of diluted solution;
M1V1 = M2V2
Rearranging the formula;
V2 = M1V1 ÷ M2
= (1.00 M × 50 mL) ÷ 0.25 M
= 200 mL
Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.
The volume of the dry gas at stp is calculated as follows
calculate the number on moles by use of PV =nRT where n is the number of moles
n is therefore = Pv/RT
P = 0.930 atm
R(gas contant= 0.0821 L.atm/k.mol
V= 93ml to liters = 93/1000= 0.093L
T= 10 + 273.15 = 283.15k
n= (0.930 x0.093) /(0.0821 x283.15) = 3. 72 x10^-3 moles
At STp 1 mole = 22.4L
what about 3.72 x10^-3 moles
by cross multiplication
volume = (3.72 x10^-3)mole x 22.4L/ 1 moles = 0.083 L or 83.3 Ml