LD50 is defined as the lethal dose 50% which describes the amount of material required to kill 50% of the testing population. It is given in units of mg of chemical per kg of bodyweight of the recipient.
Comparing hydrogen peroxide and acetic acid, we see that peroxide has a lower LD50 of 900 mg/kg, with acetic acid having LD50 = 3310 mg/kg. When comparing LD50 values, the smaller value will be the more toxic compound. What this means is that in this case, a smaller amount of peroxide is required to kill 50% of the testing population compared to acetic acid.
Therefore, 3% hydrogen peroxide is more hazardous to consume.
Answer:
1.0190 x 10⁻⁵ mol
Explanation:
We know the titration required 10.19 mL of 0.001000 M KIO₃, from this information we can calculate the number of moles KIO₃ reacted and from there the number of moles of ascorbic acid since it is a monoprotic acid ( 1 equivalent of ascorbic acid to one equivalent KIO₃).
Molarity = mol/V
V KIO₃ = 10.19 mL = 10.19 mL x 1 L/1000 mL = 0.01019 L
⇒ mol KIO₃ = V x M = 0.01019 L x 0.0010 mol / L = 1.0190 x 10⁻⁵ mol KIO₃
# mol ascorbic acid = # mol KIO₃ = 1.0190 x 10⁻⁵ mol
Answer:
0.85 mole
Explanation:
Step 1:
The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:
When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
Step 2:
Conversion of 85g of CaCO3 to mole. This is illustrated below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 = 85g
Moles of CaCO3 =?
Number of mole = Mass /Molar Mass
Mole of CaCO3 = 85/100
Mole of caco= 0.85 mole
Step 3:
Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.
This is illustrated below :
From the balanced equation above,
1 mole of CaCl2 reacted to produced 1 mole of CaCO3.
Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.
From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3
Answer:
carbondioxide is an example of compound.
Answer:
45.02 L.
Explanation:
- Firstly, we need to calculate the no. of moles of water vapor.
- n = mass / molar mass = (36.21 g) / (18.0 g/mol) = 2.01 mol.
- We can calculate the volume of knowing that 1.0 mole of a gas at STP occupies 22.4 L.
<em><u>Using cross multiplication:</u></em>
1.0 mole of CO occupies → 22.4 L.
2.01 mole of CO occupies → ??? L.
∴ The volume of water vapor in 36.21 g = (22.4 L)(2.01 mole) / (1.0 mole) = 45.02 L.
