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Leno4ka [110]
3 years ago
12

Does acid in your stomach make you unhealthy?

Chemistry
2 answers:
Delvig [45]3 years ago
8 0
No because it burns the food that u had and leaves it more space?????? I
Umnica [9.8K]3 years ago
7 0

This can happen when eating large amounts of any food, not just foods known to trigger your heartburn symptoms. Fatty foods are big no-nos if you suffer from heartburn. High-fat foods sit around in your belly longer. This makes your stomach produce more acid, irritatingyour digestive system.

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Pls help!!! ILL GIVE BRAINLIEST, 70 points + 10 more if not plagiarized ty ty (also for those who scam and do it for the points
sukhopar [10]

Answer:

What procedure did you use to complete the lab?

→Procedures that needs to be considered to complete the lab are- a thorough knowledge of lab assignments, knowledge about safety equipment, reviewing the MSDS of chemicals for lab experiment etc.

<h3>Explanation:</h3>

To be lab prepared one must follow these procedures-

1. One should have the knowledge of lab assignments to make the lab experiment easier.

2. To be aware about safety equipment and their uses in lab, like- the location of fire extinguisher in lab.

3. To know the steps of experiments to be performed

4. To fill notebook of lab with information regarding the experiment

5. One should review the data sheets of chemicals material safety.

6. To put on all the necessary dressings to perform experiment.

7. To have complete understanding about the experiment disposals.

7 0
2 years ago
Read 2 more answers
Joan needs a 0.053-M solution of HCl, but only has access to 2.12-M HCl, a 10-mL graduated pipet, and a 25-mL volumetric flask.
dmitriy555 [2]

Answer:

Joan should measure 0.625 mL of 2,12 M HCl solution to create her desired solution.

Explanation:

Molarity of HCl Joan has access to = M_1=2.12 M

Volume of 2.12 M of HCl Joan use = V_1=?

Molarity of HCl Joan desired = M_2=0.053 M

Volume of 0.053 M of HCl Joan can prepare = V_2=25 mL

M_1V_1=M_2V_2

V_1=\frac{M_2V_2}{M_1}

=\frac{0.053 M\times 25 mL}{2.12 M}=0.625 mL

Joan should measure 0.625 mL of 2,12 M HCl solution to create her desired solution.

6 0
3 years ago
When mantle rocks near the radioactive core are heated, they become less dense than the cooler, upper mantle rocks. These warmer
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C) convection currents
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3 years ago
In a chemical equation, the sum of the masses of the products is equal to the sum of the masses of the reactants. True False
Komok [63]
The answer is true e e e e e  e e e e e
8 0
3 years ago
Read 2 more answers
Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m
olga nikolaevna [1]

Explanation:

The reaction equation for the given reaction will be as follows.

               HCl + NaOH \rightarrow NaCl + H_{2}O

Each mole of both HCl and NaOH gives one mole of water.

Also, it is given that 1 liter of NaOH and HCl solution contains 1.6 mol dm^{-3} of NaOH (HCl).

It is known that 1 cm^{3} = 0.001 liter. So, 87 cm^{3} = 0.087 liter.

Hence, number of moles of water obtained from the given reaction are as follows.

                  0.087 liter × 1.6 mol = 0.1392 moles

            No. of moles = \frac{mass}{molar mass of water}

                0.1392 moles = \frac{mass}{18 g/mol}

                       mass = 2.5056 g

Now, volume of water present before the reaction is 2 \times 0.087 liter = 0.174 liter or 0.174 Kg (as density is 1 Kg/cm^{3}) or 174 g (as 1 kg = 1000 g).

Therefore, total weight of water present = 2.5056 g + 174 g = 176.5056 g

Formula to calculate enthalpy of neutralization is as follows.

                Enthalpy of neutralization = mS \Delta T

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where,            m = mass

                       S = specific heat capacity

                   \Delta T = change in temperature

Putting the given values in the formula as follows.

       Enthalpy of neutralization = mS \Delta T

                     = 176.5056 g \times 4.18 J/K g \times (317.4 K - 298 K)              

                                                           = 14333.73 J

   or,                                                    = 14.33 kJ

Thus, we can conclude that the enthalpy of neutralization of given reaction is 14.33 kJ.

5 0
4 years ago
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