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OleMash [197]
3 years ago
11

If someone is suffering from the problem of acidity after overeating, which of the following would you suggest as remedy ?

Chemistry
1 answer:
kolezko [41]3 years ago
6 0

Answer:

nth

Explanation:

You might be interested in
Which substance is correctly identified as a homogeneous or heterogeneous mixture?
larisa86 [58]

Answer:

A

Explanation:

A salad with lettuce, cheese, seeds, tomatoes, broccoli, and other vegetables is an example of a heterogeneous mixture.

Milk is not mixed with anything so that is incorrect

River water is mixed in with dirt and minerals so this one is incorrect

Beach sand is incorrect because you can see colored particles.

7 0
3 years ago
What volume of nitrogen dioxide is formed at 735 torr and 28.2 °C by reacting 3.56 cm3 of copper (d = 8.95 g/cm3) with 200 mL of
weqwewe [10]

Answer:

25.76 L

Explanation:

Given, Volume of Copper = 3.56 cm³

Density = 8.95 g/cm³

Considering the expression for density as:

Density=\frac {Mass}{Volume}

So,

So, Mass= Density * Volume = 8.95 g/cm³ * 3.56 cm³ = 31.862 g

Mass of copper = 31.862 g

Molar mass of copper = 63.546 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{31.862\ g}{63.546\ g/mol}

<u>Moles of copper = 0.5014 moles </u>

Given, Volume of nitric acid solution = 200 mL = 200 cm³

Density = 1.42 g/cm³

Considering the expression for density as:

Density=\frac {Mass}{Volume}

So,

So, Mass= Density * Volume = 1.42 g/cm³ * 200 cm³ = 284 g

Also, Nitric acid is 68.0 % by mass. So,  

Mass of nitric acid = \frac {68}{100}\times 284\ g = 193.12 g

Molar mass of nitric acid = 63.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{193.12\ g}{63.01\ g/mol}

<u>Moles of nitric acid = 3.0649 moles </u>

According to the reaction,  

Cu_{(s)}+4HNO_3_{(aq)}\rightarrow Cu(NO_3)_2_{(aq)} + 2NO_2_{(g)} + 2H_2O_{(l)}

1 mole of copper react with 4 moles of nitric acid

Thus,  

0.5014 moles of copper react with 4*0.5014 moles of nitric acid

Moles of nitric acid required = 2.0056 moles

Available moles of nitric acid = 3.0649 moles

<u>Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent. </u>

The formation of the product is governed by the limiting reagent. So,

1 mole of copper on reaction forms 2 moles of nitrogen dioxide

So,

0.5014 mole of copper on reaction forms 2*0.5014 moles of nitrogen dioxide

<u>Moles of nitrogen dioxide = 1.0028 moles </u>

Given:  

Pressure = 735 torr

The conversion of P(torr) to P(atm) is shown below:

P(torr)=\frac {1}{760}\times P(atm)

So,  

Pressure = 735 / 760 atm = 0.9632 atm

Temperature = 28.2 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28.2 + 273.15) K = 301.35 K  

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9632 atm × V = 1.0028 mol × 0.0821 L.atm/K.mol × 301.35 K  

<u>⇒V = 25.76 L</u>

4 0
3 years ago
Calculate the volume of carbon dioxide and water vapour produced and the volume of oxygen remaining, when 20.0 dm3 of propane re
laiz [17]

Answer:

80cm3 of water, and 60cm3 carbon IV oxide is formed while 20cm3 of oxygen is left unreacted.

Explanation:

From Gay-Lussac's law, there are five volumes of oxygen, 1 volume if propane, 4 volumes of water and three volumes of CO2. Applying this shows the reacting volumes as we have in the image attached, hence the volumes left after reaction.

6 0
3 years ago
Which of the substances are elements please help
ivanzaharov [21]

Answer:

Substances 1 and 2

Explanation:

an element only has 1 kind of atoms :3

6 0
3 years ago
Describe what golgi looks like.
Margaret [11]
Here is what Golgi looks like, so that you can look at the picture and describe it.
______________________________________________________________
____________________________________________________________________________________________________________________________

Glad I could help, and good luck!
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7 0
3 years ago
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