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Scorpion4ik [409]
3 years ago
9

When a white powder is added to a liquid in a beaker at room temperature, the beaker becomes too hot to touch. What kind of reac

tion is this an example of?
A. a fusion reaction
B. a combustion reaction
C. an exothermic reaction
D. an endothermic reaction
Chemistry
2 answers:
Varvara68 [4.7K]3 years ago
6 0
<span>C. an exothermic reaction
</span>
liberstina [14]3 years ago
3 0
The answer is <span>an C.exothermic reaction</span>
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3 years ago
Calculate the pH of a solution in which one normal adult dose of aspirin (640 mg ) is dissolved in 10 ounces of water. Express y
d1i1m1o1n [39]

The pH of the solution in which one normal adult dose aspirin is dissolved is :  2.7

Given data :

mass of aspirin = 640 mg = 0.640 g

volume of water = 10 ounces = 0.295735 L

molar mass of aspirin = 180.16 g/mol

moles of aspirin = mass / molar mass = 0.00355 mol

<h3>Determine the pH of the solution </h3>

First step : <u>calculate the concentration of aspirin</u>

= moles of Aspirin / volume of water

= 0.00355 / 0.295735

= 0.012 M

Given that pKa of Aspirin = 3.5

pKa = -logKa

therefore ; Ka = 10^{-3.5} = 3.162 * 10^{-4}

From the Ice table

3.162 * 10^{-4} = \frac{x + H^+}{[aspirin]}  = \frac{x^{2} }{0.012-x}

given that the value of Ka is small we will ignore -x

x² = 3.162 * 10^{-4} * 0.012

x = 1.948 * 10^{-3}  

Therefore

[ H⁺ ] = 1.948 * 10^{-3}

given that

pH = - Log [ H⁺ ]

     = - ( -3 + log 1.948 )

     = 2.71 ≈ 2.7

Hence we can conclude that The pH of the solution in which one normal adult dose aspirin is dissolved is :  2.7

Learn more about Aspirin : brainly.com/question/2070753

4 0
2 years ago
A colorless liquid has a molar mass of 60.01 g/mol. When the liquid was analyzed, it was 46.7% nitrogen and 53.3% oxygen. What i
MAXImum [283]

first we have to find the empirical formula of the compound

empirical formula is the simplest ratio of whole numbers of components making up a compound

for 100 g of the compound

N O

mass 46.7 g 53.3 g

number of 46.7 g/ 14 g/mol 53.3 g/ 16 g/mol

moles = 3.34 mol = 3.33 mol

divide by the least number of moles

3.34/3.33 = 1.00 3.33/ 3,33 = 1.00

therefore number of atoms are

N - 1

O - 1

empirical formula is - NO


mass of empirical unit - 14 g/mol + 16 g/mol = 30 g

molecular formula is actual composition of elements in the compound

molecular mass - 60.01 g/mol


number of empirical units = molecular mass / empirical unit mass

= 60.01 g/mol / 30 g = 2

there are 2 empirical units


2(NO)

molecular formula = N₂O₂


6 0
3 years ago
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