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1 year ago

What is a common use of bases?

Chemistry

1 answer:

Fittoniya [83]1 year ago

3 0

A common use of bases is to reduce indigestion. Hence, option D is correct.

A base is a substance that can neutralize the acid by reacting with hydrogen ions.

Bases are substances whose solutions have a pH value of more than 7. Bases are bitter in taste.

Whereas acids are the substances whose solutions have a pH value of less than 7. Acids are sour in taste.

When an excess of acid is formed in the human stomach then an antacid tablet which is basically a base is consumed that helps in reducing indigestion.

Therefore, we can conclude that the common use of bases is to reduce indigestion.

Learn more about the base here:

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Which formula represents the compound commonly known as phosphine

Arlecino [84]

I looked up what is the molecular formula for Phosphine and got this: PH3
Hope this helps! Let my know if this was correct.

8 0

3 years ago

WS Percent yield don’t understand how to do would appreciate the help

Grace [21]

Answer:

1. Theoretical yield of NaOH is 22.72 g

2. Percentage yield of NaOH = 22.14%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NaHCO₃ —> NaOH + CO₂

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 1 mole (i.e 40 g) of NaOH and 1 mole (i.e 44.01 g) of CO₂.

Next, we shall determine the number of mole of NaHCO₃ that will decompose to produce 25 g of CO₂. This can be obtained as follow:

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 44.01 g of CO₂.

Therefore, Xmol of NaHCO₃ will decompose to 25 g of CO₂ i.e

Xmol of NaHCO₃ = 25 / 44.01

Xmol of NaHCO₃ = 0.568 mole

1. Determination of the theoretical yield of NaOH.

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 40 g of NaOH.

Therefore, 0.568 mole of NaHCO₃ will decompose to produce = 0.568 × 40 = 22.72 g of NaOH.

Thus, the theoretical yield of NaOH is 22.72 g

2. Determination of the percentage yield of NaOH.

Theoretical yield of NaOH = 22.72 g

Actual yield of NaOH = 5.03 g

Percentage yield of NaOH =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 5.03 / 22.72 × 100

Percentage yield of NaOH = 22.14%

4 0

2 years ago

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

4 0

3 years ago

How many significant figures does the number have?<br> 9,000,100

Sergeeva-Olga [200]

Answer:

5 Significant Figures

Explanation:

6 0

3 years ago

What is the mass of a 0.5 mol sample of C2H6?

Eddi Din [679]

Answer: 30.08 g/mol

Explanation:

C2= 12.01 H6=1.01

12.01(2) + 1.01(6) =

24.02 + 6.06 =

30.08 g/mol

3 0

2 years ago