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Mumz [18]
4 years ago
9

What name is given to substances that resist changes in ph?

Chemistry
1 answer:
nignag [31]4 years ago
4 0
These are buffer systems (solutions of substances) with a steady concentration of hydrogenium ions. At addition to such solution of a small amount of acid or alkali does not change рН.
For example:
CH₃COOH + CH₃COONa    acetate buffer
KH₂PO₄ + K₂HPO₄  potassium phosphate buffer
You might be interested in
H3C6H507(aq) + 3NaHCO3(aq) - Na3C6H5O7(aq) + 3C02(g) + 3__
zavuch27 [327]

Answer: 3<u> H₂O</u>

<u />

Explanation:

This is an example of a common chemical reaction

ACID + CARBONATE --> SALT + WATER + CARBON DIOXIDE which implires

CITRIC ACID  + SODIUM BICARBONATE ----> SODIUM CITRATE + WATER + CARBON DIOXIDE .  

The full chemical equation is  given by

H3C6H5O7(aq)  + 3NaHCO3(aq) --> Na3C6H5O7(aq) + 3CO2(g) + 3<u>H2O</u>(l)

The reaction produces a  fizz due to the  release of  carbon dioxide which can be also seem when an Alka-Seltzer® tablet is dissolved in water.

6 0
4 years ago
Read 2 more answers
What is the smallest part of an element that still retains properties of that element?
Leni [432]
The smallest part is the atom
4 0
3 years ago
Read 2 more answers
A 0.568-g sample of fertilizer contained nitrogen as ammonium sulfate, . It was analyzed for nitrogen by heating with sodium hyd
Mariulka [41]

Answer:

6.69%

Explanation:

Given that:

Mass of  the fertilizer = 0.568 g

The mass of HCl used in titration (45.2 mL of 0.192 M)

= 0.192*\frac{45.2}{1000}* \frac{36.5}{1 \ mole}

= 0.313 g HCl

The amount of NaOH used for the titration of excess HCl (42.4 mL of 0.133M)

= \frac{44.3 \ mL * 1.0 \ L}{1000 \ mL} *0.133 \ mole/L

= 0.0058919 mole of NaOH

From the neutralization reaction; number of moles of HCl is equal with the number of moles of NaOH is needed to complete the neutralization process

Thus; amount of HCl neutralized by 0.0058919 mole of NaOH = 0.0058919 × 36.5 g

= 0.2151 g HCl

From above ; the total amount of HCl used = 0.313 g

The total amount that is used for complete neutralization = 0.2151 g

∴ The amount of HCl reacted with NH₃ = (0.313 - 0.2151)g

= 0.0979 g

We all know that 1 mole of NH₃ = 17.0 g , which requires 1 mole of HCl = 36.5 g

Now; the amount of HCl neutralized by 0.0979 HCl = \frac{17}{36.5}*0.0979

= 0.0456 g

Therefore, the mass of nitrogen present in the fertilizer is:

= 0.0456 \ g \ NH_3 * \frac{1  \ mol \ NH_3 }{17.0 \ mol \ of \ NH_3} * \frac{1  \mol \ (NH_4)_2SO_4}{2 \ mol \ NH_3 } * \frac{2 \ mol \ N }{1  \mol \ (NH_4)_2SO_4}* \frac{14.0 g }{1 \ mol \ N}

= 0.038 g

∴ Mass percentage of Nitrogen in the fertilizer = \frac{0.038 \ g}{0.568 \ g} * 100%

= 6.69%

8 0
4 years ago
Read 2 more answers
An atom has9 electrons and 9 protons at the start. If it loses 2 electrons, what would the net charge on the atom be
nevsk [136]

Answer:

i hope it will help you

Explanation:

there will be 7 electrons and 9 protons will have 2+ charge.

3 0
3 years ago
A textbook measures 250 mm long, 225 mm wide and 50 mm thick. What is the volume of this book in mm3? What is the volume of this
Dvinal [7]

Answer:

2.81 × 10⁶ mm³

2.81 × 10⁻³ m³

Explanation:

Step 1: Given data

Length (l): 250 mm

Width (w): 225 mm

Thickness (t): 50 mm

Step 2: Calculate the volume of the textbook

The book is a cuboid so we can find its volume (V) using the following expression.

V = l × w × t = 250 mm × 225 mm × 50 mm = 2.81 × 10⁶ mm³

Step 3: Convert the volume to cubic meters

We will use the relationship 1 m³ = 10⁹ mm³.

2.81 × 10⁶ mm³ × 1 m³ / 10⁹ mm³ = 2.81 × 10⁻³ m³

6 0
3 years ago
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