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sergeinik [125]
3 years ago
9

aspirin C6H4(CO2)(CO2CH3),can be prepared in the chemistry laboratory by the reactions of salicylic acid, C6H4(CO2H)(OH),with ac

etic anhydride(CH3CO)2O.what volume of acetic anhydride(density,1.0820g/cm3) is required to produce 1.00kg of aspirin, assuming a 100% yield

Chemistry
1 answer:
Ket [755]3 years ago
8 0
262mol 1=kg
g=1000
from stoichoimetry

x=102*1000/360
x=102000/360
x=283.33

density =m/v
=283.33/1.082
=262mol

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cluponka [151]
See the correct answer is b.)PbS
8 0
3 years ago
You bought a new car and estimated that your monthly payment would be $312. However, your actual payment amount was $325. How mu
Romashka-Z-Leto [24]

Answer:

The error would be $13.

Explanation:

Given data:

Actual payment = $325

estimated payment = $312

Error = ?

Solution:

Error =  Estimated payment - Actual payment

Error = $312  -  $325

Error = -$13

we can discard the negative sign and consider the absolute value. The error would be $13.

5 0
3 years ago
Acids and bases can neutralize each other in double displacement reactions. For example, if hydrogen chloride (a strong acid, HC
scoray [572]

Answer:

The answer to your question is    HCl  +  NaOH   ⇒   NaCl  +  H₂O

Explanation:

Data

Double displacement reaction

Balanced chemical reaction

               HCl  +  NaOH   ⇒   NaCl  +  H₂O

           Reactants           Elements                Products

                 1                  Chlorine (Cl)                    1

                 1                   Sodium (Na)                   1

                 2                 Hydrogen (H)                  2

                 1                   Oxygen (0)                      1

As we can see, the reaction is balanced and the coefficients of all reactants and products are 1, but the number is not written in a balanced reaction.

8 0
3 years ago
10. What base unit of measurement is common to all systems?
Lera25 [3.4K]

Answer:

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5 0
2 years ago
To determine the ammonia concentration in a sample of lake water, three samples are prepared. In sample A, 10.0 mL of lake water
Masteriza [31]

Answer:Sample Absorbance (625 nm)  

A 0.536  

B 0.783  

C 0.045  

Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.

First, we see 1 mole of NH3 gives 1 mole product.

In B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of NH3 in A) vol of B = 25 = vol of A

now A = el C = eC ( since l = 1cm)

Because, n net absorbance due to complex blank absorbance must be removed.

Here A(A) = 0.536 - 0.045 = 0.491 , A(B) = 0.783 - 0.045 = 0.738  

(you can plug in different numbers in this step)

A2/A1 = C2/C1 , A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491

So, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)

Hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

Lake water vol = 10 ml out of 25,

Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

Then, A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1

Explanation:

4 0
3 years ago
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