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netineya [11]
4 years ago
6

Aspirin is the common name for acetyl salicylic acid, CgH804. One tablet contains 0.325 g of aspirin. How many moles of aspirin

is
this?
Chemistry
2 answers:
d1i1m1o1n [39]4 years ago
7 0

Answer:

0.0018 moles

Explanation:

Given data:

mass of aspirin= 0.325 g

molar mass of aspirin= 180.1589 g/mol

number of moles=?

Formula:

number of moles= mass in gram/ molar mass (g/mol)

Solution:

now we will put the values in given formula:

number of moles of aspirin= mass of aspirin/ molar mass of aspirin

number of moles= 0.325 g/ 180.1589 g/mol

number of moles = 0.0018 mol

Mole:

Mole is the unit of amount in chemistry. it is SI unit and represented by "mol".

Example;

one mole of hydrogen has a mass of 1.008 g and contain 6.022× 10∧23  atoms.

vazorg [7]4 years ago
7 0

Answer:

58.5 moles acetyl salicylic acid

Explanation:

Molar mass of acetyl salicylic acid, C9H804

C= 12g/mole ⇒ 9*12 = 108g/mole

H=1g/mole ⇒ 8*1 =8

O=16g/mole ⇒ 4*16 =64g/mole

⇒108+8+64 = 180g/mole

Moles acetyl salicylic acid:

mole = mass * Molar mass

moles acetyl salicylic acid = 0.325g * 180g/mole = 58.5 moles acetyl salicylic acid

This aspirin, of 0.325g, has 58.5 moles of acetyl salicylic acid

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what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid require
vivado [14]

Answer:

pH at equivalence point is 8.52

Explanation:

HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of HCOO^{-}

So, moles of NaOH used to reach equivalence point equal to number of moles HCOO^{-} produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of HCOO^{-} produced = \frac{29.80\times 0.3567}{1000}moles=0.01063moles

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of HCOO^{-} = \frac{0.01063\times 1000}{54.80}M=0.1940M

At equivalence point, pH depends upon hydrolysis of HCOO^{-}. So, we have to construct an ICE table.

HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}

I: 0.1940                                   0                 0

C: -x                                          +x               +x

E: 0.1940-x                                x                  x

So, \frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}

species inside third bracket represent equilibrium concentrations

So, \frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}

or,x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0

So, x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}

So, x=3.285\times 10^{-6}M

So, pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52

5 0
3 years ago
7th grade SCIENCE PLEASE HELP ​
Flura [38]

Answer:  the first one is adaptation, third is migration, 4th is hibernation,

Explanation:

i dont know the other ones, sorry because i am also a 7th grader.

4 0
3 years ago
Which metal will lose electrons more easily, Ca or Mg?
alisha [4.7K]
Yes it is not or it will
3 0
3 years ago
A certain pressure of n2o4 is initially added to an otherwise evacuated rigid vessel. at equilibrium, 25.8% of n2o4 remains. det
mote1985 [20]
The pressure is 902 x10 as it has partial adequates which is linguistic in an average point
6 0
3 years ago
A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The
natima [27]
1) Compund Ir (x) O(y)

2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g

3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g

4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g

5) Convert grams to moles

moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles

moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131

6) Find the proportion of moles

Divide by the least of the number of moles, i.e. 0.00656

Ir: 0.00656 / 0.00656 = 1

O: 0.0131 / 0.00656 = 2

=> Empirical formula = Ir O2 (where 2 is the superscript for O)

Answer: Ir O2
5 0
4 years ago
Read 2 more answers
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