Answer:
Bromobenzoic acid
Explanation:
COOH has higher priority than Br, so the parent name is benzoic acid.
The COOH group is automatically on C1 of the ring, and the other aromatic carbons are numbered sequentially around the ring.
The name of the substituent is joined to that of the parent, and the locant is attached to that with a hyphen.
The generic IUPAC name is bromobenzoic acid.
There are three isomers of benzoic acid — 2-bromobenzoic acid, 3-bromobenzoic acid, and 4-bromobenzoic acid — so you must always specify which one you mean.
Stoichiometry:
First, calculate the number of grams for one mole of Ca3 (PO3)4
(3 * (Mass of Ca)) + (4 * (Mass of P + (3 * Mass of Oxygen)))
= (3*40.08) + 4(30.97 + (3*16.00))
=(120.24) + 4(78.97)
=436.12 g / mol Ca3(PO3)4
This means there are 436.12 g per 1 mole of Ca(PO3)4. Since there are 4.50 moles of Calcium Phosphate, mulitply the molar mass of Ca(PO3)4 by 4.50 and you should get 1962.54 g. Since there are 3 sigfigs, the final answer is 1960 g.
on a side note: I put in all my work in case 1. your periodic table if different, 2. my work is wrong, 3. you put in the question wrong because I feel that the actual compound would be Ca3(PO4)3 instead of Ca3(PO3)4 (if this is the case, the answer should be 1820 g).
Mg + 2AgCl —> MgCl2 + 2Ag
Answer: 52.4 grams
Explanation:
The two reactants PCl5 and H2O, must occur in a 1:4 ratio, so that tells us, that we need 4 times more H₂O than phosphorus pentachloride.
0.287 moles PCl5 multiplied by 4 equal 1.15 moles H2O. That tell sus that 1.15 moles H2O is needed.
Because we have 3.84 moles H2O, the PCL5 is the limiting reactant.
0.287 mol PCl₅ x (5 mol HCl divided by 1 mol of PCl₅) x ( 36.5g HCl divided by moles of HCl) = 52.4 g of HCl is formed
(Hopefully this helps!)
Sample means for solutions 1 and 2 are 19.27 and 10.32 respectively
In semiconductor manufacturing,
The total for answer 1 is given by:
9.7+10.5+9.4+10.6+9.3+10.7+9.6+10.4+10.2+10.5 = 192.7
The sample size is 10 and provides us with
192.7/10 = 19.27
For solution 2, the sum is given by:
10.6+10.3+10.3+10.2+10.0+10.7+10.3+10.4+10.1+10.3 = 103.2
The sample size is 10, this gives us
103.2/10 = 10.32
The total for answer 2 is given by:
10.6+10.3+10.3+10.2+10.0+10.7+10.3+10.4+10.1+10.3 = 103.2
The sample size is 10 and provides us with
103.2/10 = 10.32
Learn more about semiconductor manufacturing here brainly.com/question/22779437
#SPJ4.
In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etch rate is an important characteristic in this process and is known to follow a normal distribution. Two different etching solutions have been compared, using two random samples of 10 wafers for each solution. Assume the variances are equal. The etch rates are as follows (in mils per minute): Solution 1 Solution 2 9.7 10.6 10.5 10.3 9.4 10.3 10.6 10.2 9.3 10.0 10.7 10.7 9.6 10.3 10.4 10.4 10.2 10.1 10.5 10.3 Calculate sample means of solution 1 and solution 2
