Answer:
Sodium hydroxide is <u>very corrosive</u>. It can cause irritation to the eyes, skin, and mucous membrane; an allergic reaction; eye and skin burns; and temporary loss of hair. Workers may be harmed from exposure to sodium hydroxide.
Explanation:
<u>How to reduce:</u>
Inhalation: Move victim to fresh air.
Skin Contact: Avoid direct contact
Eye Contact: Avoid direct contact
Ingestion: Have victim rinse mouth with water.
Answer:
The abundance of first isotope is 69.15 %
The abundance of second isotope is 30.85 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope:
% = x %
Mass = 62.9296 u
For second isotope:
% = 100 - x
Mass = 64.9278 u
Given, Average Mass = 63.546 u
Thus,
Solving for x, we get that:
x = 69.15 %
<u>The abundance of first isotope is 69.15 %</u>
<u>The abundance of second isotope is 100 - 69.15 % = 30.85 %</u>
Answer:
240 g/mol
Explanation:
We'll begin by writing the chemical formula of Disodium sulfide nonahydrate. This is illustrated below:
The molecular formula of disodium sulfide nonahydrate is Na₂S.9H₂O
Finally, we shall determine the molar mass of disodium sulfide nonahydrate, Na₂S.9H₂O. This can be obtained by adding the individual molar mass of the element present in the compound as illustrated below:
Molar mass of Na₂S.9H₂O = (23×2) + 32 + 9[(2×1) + 16]
= 46 + 32 + 9[2 + 16]
= 46 + 32 + 9[18]
= 46 + 32 + 162
= 240 g/mol
Thus, the molar mass of disodium sulfide nonahydrate, Na₂S.9H₂O is 240 g/mol
The H₂O is the formula of the chemical specie while the (s), (l) and (g) tell us the state in which it is present, where s is solid, l is liquid and g is gas.
Answer:
M of HI = 5.4 M.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
<em>(XMV) acid = (XMV) base.</em>
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HI = (XMV) Ca(OH)₂.</em>
For HI; X = 1, M = ??? M, V = 25.0 mL.
For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.
<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>
