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4 years ago

When a beverage is “carbonated” what is happening? How does it affect the acidity?

Chemistry

1 answer:

Ksju [112]4 years ago

8 0

Answer:

In addition, carbonated beverages may serve as an acid load and thus may raise gastric acid volume, leading to increased likelihood of gastro‐oesophageal reflux. Two studies have demonstrated that carbonated beverages can reduce the oesophageal pH < 4 and potentially cause GERD‐related symptoms.

Explanation:

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Determine the freezing point depression of 2 kg of water when 2 mol salt is added to it. The kf of water is 1.86 degrees C/M. Sh

Svet_ta [14]

Answer: The freezing point depression is $1.86^0C$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=K_f\times m$

$\Delta T_f$ = Depression in freezing point

$K_f$ = freezing point constant = $1.86^0C/m$

m= molality = $\frac{\text {moles of solute}}{\text {mass of solvent in kg}}=\frac{2mol}{2kg}=1m$

$\Delta T_f=1.86mol/kg^0C\times 1m$

$\Delta T_f=1.86^0C$

Thus freezing point depression is $1.86^0C$

3 0

3 years ago

Actual conjugate acid of acetamide

viva [34]

CH3-C-NH2

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4 years ago

Eukaryotic gene expression is regulated during transcription and RNA processing, which take place in the nucleus, and during pro

Arlecino [84]

Answer:

Ok, is this a true and false question? Anyways, yes. the central dogma, from dna transcription to mRNA translation to proteins. and further regulation can occur through post-translational modifications of proteins. it is a true statement

Explanation:

6 0

3 years ago

Your dad is working on creating a brick border for the lake in your backyard. Each brick has a mass of 100 g and a volume of 20

beks73 [17]

Density=mass/volume=100g/20cm^3=5g/cm^3. The brick will fall in the lake, because its density is higher than that of water (1g/cm^3).

4 0

3 years ago

To begin the experiment, 1.5g of methane CH4 is burned in a bomb calorimeter containing 1,000 grams of water. The initial temper

vredina [299]

Mass of methane takne = 1.5g

moles of methane used = masss / molar mass = 1.5 / 16 = 0.094 moles

mass of water = 1000 g

Initial temperature of water = 25 C

final temperature = 37 C

specific heat of water = 4.184 J /g C

1) Heat absorbed by water = q =m• C• ΔT = 1000 X 4.184 x (37-25) = 50208 Joules

2) Heat absorbed by calorimeter = Heat capacity X ΔT = 695 X (37-25) = 8340 J

3) Total heat of combustion = heat absorbed by water + calorimeter = 50208 + 8340 = 58548 Joules

This heat is released by 0.094 moles of methane

So heat released by one mole of methane =

- 622851.06 Joules = 622.85 kJ / mole

4) standard enthalpy of combustion = -882 kJ / mole

Error = (882-622.85) X 100 / 882 = 24.84 %

4 0

3 years ago