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docker41 [41]
3 years ago
11

A science student observes that a dissolvable antacid tablet acts faster in relieving stomach pain when taken with warm water th

an when taken with water at room temperature. Based on the observation, suggest a potential problem that can be studied. Also, list the dependent, independent, and controlled variables that can exist in such an investigation.
Chemistry
1 answer:
Alika [10]3 years ago
6 0

Answer:

We know that an antacid is a base that can neutralize the stomach acid when ingested. The student observed that a soluble antacid tablet reacted faster in relieving pain due to the excess acid secreted in the stomach. So, the problem that can be studied is the effect of temperature on the rate of neutralization of stomach acid by the antacid. The independent variable is the one the student is going to vary that is the temperature. The dependent variable is the rate of reaction or the time taken for an antacid tablet to neutralize the acid present in stomach. The controlled variable is the tablet taken, that is the student should use a specific antacid tablet to ensure that the rate of reaction or the time taken for the antacid tablet to neutralize the acid present in stomach is not affected by the chemical composition of tablet.

Explanation:

Hope u understood.

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What is the pressure of 1.50 moles of a gas in a 30.0L tank at a temperature of<br> 285K?
julia-pushkina [17]

The pressure of the gas : 1.1685 atm

<h3>Further explanation</h3>

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.08206 L.atm / mol K  

T = temperature, Kelvin  

n=moles=1.5

V=volumes = 30 L

T=temperature=285 K

The pressure :

\tt P=\dfrac{nRT}{V}\\\\n=\dfrac{1.5\times 0.082\times 285}{30}\\\\P=1.1685~atm

7 0
3 years ago
The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0°c is 4.48. what is the value of ka for hbro?
elena-14-01-66 [18.8K]

The solution would be like this for this specific problem:

Given:

 

pH of a 0.55 M hypobromous acid (HBrO) at 25.0 °C =  4.48

 

[H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-] <span>

Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9</span>

 

To add, Hypobromous Acid does not require acid adjustment, which is necessary for chlorine-based product and is stable and effective in pH ranges of 5-9.<span>

</span>Hypobromous Acid combines with organic compounds to form a bromamine. Chlorine also combines with the same organic compounds to form a chloramine. <span>It is also one of the least expensive intervention antimicrobial compounds available.</span>

8 0
4 years ago
Read 2 more answers
Which option is a mixture?<br> A. potassium<br> B. sugar<br> C. water<br> D. salad
PSYCHO15rus [73]

Answer:

D.Salad

Explanation:

A mixture is a substance that combines two or more substance that are physically combined. A mixture can be easily seperated by a physical means . A mixture is the combination of various substances that are not chemically combined .Examples of mixtures are cake and salad . The  cake contains various substances like egg, flour and sugar. This mixture can be easily separated into it individual entities by a physical means .

Base on the option the only mixture among the option is salad . Salad is a a mixtures that combines substances like tomatoes , fruits and vegetable. The individual substances that makes up the salad can be easily separated.

Sugar is a compound . Water is a compound as it contains two elements and potassium is an element.

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3 years ago
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Why it is advisided to freshly prepare the ferrous sulphate solution for experiment
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Here is a site my buddie has to help you. Well co-owner..
https://www.quora.com/Why-is-fresly-prepared-FeSO4-required-for-the-ring-test
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3 years ago
2. Calculate the atomic mass of an element that has two isotopes, each with 50.00% abundance. One isotope has a mass of 63.00 am
melamori03 [73]

Answer:

The atomic mass of element is 65.5 amu.

Explanation:

Given data:

Abundance of X-63 = 50.000%

Atomic mass of  X-63 = 63.00 amu

Atomic mass of X-68 = 68.00 amu

Atomic mass of element = ?

Solution:

Abundance of X-68 = 100-50 = 50%

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (50×63)+(50×68) /100

Average atomic mass =  3150 + 3400 / 100

Average atomic mass  = 6550 / 100

Average atomic mass = 65.5 amu.

The atomic mass of element is 65.5 amu.

7 0
3 years ago
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