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SOVA2 [1]
3 years ago
12

How to make a explosion at home with normal stuff​

Chemistry
1 answer:
navik [9.2K]3 years ago
5 0

answer:

cleaning products

Explanation:

causes chemical reactions when mixed

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How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride (FM 121.135) to give a pH of 7.60 in a fina
liubo4ka [24]

Answer:

41.64mL of NaOH 0.500M must be added to obtain the desire pH

Explanation:

It is possible to find pH of a buffer by using H-H equation, thus:

pH = pka + log [A⁻] / [HA]

<em>Where [HA] is concentration of the weak acid TRIS-HCl and [A⁻] is concentration of its conjugate acid.</em>

Replacing in H-H equation:

7.60 = 8.072 + log [A⁻] / [HA]

0.3373 =  [A⁻] / [HA] <em>(1)</em>

10.0g of TRIS-HCl (Molar mass: 121.135g/mol) are:

10.0g ₓ (1mol / 121.135g) = 0.08255 moles of acid. That means moles of both the acid and conjugate base are:

[A⁻] + [HA] = 0.08255 <em>(2)</em>

Replacing (1) in (2):

0.3373 =  0.08255 - [HA] / [HA]

0.3373[HA] =  0.08255 - [HA]

1.3373[HA] = 0.08255

<em>[HA] = 0.06173 moles</em>

Thus:

[A⁻]  = 0.08255 - 0.06173 = 0.02082 moles [A⁻]

The moles of A⁻ comes from the reaction of the weak acid with NaOH, that is:

HA + NaOH → A⁻ + H₂O + K⁺

Thus, <em>you need to add 0.02082 moles of NaOH to produce 0.02082 moles of A⁻. </em>As NaOH solution is 0.500M:

0.02082 moles NaOH ₓ (1L / 0.500mol) = 0.04164L of NaOH 0.500M =

<h3>41.64mL of NaOH 0.500M must be added to obtain the desire pH</h3>

3 0
3 years ago
Select the correct electron configuration for Boron. (Atomic Number 5) 1s 22s 12p 2 1s 22s 3 1s 32s 12p 1 1s 22s 22p 1
kodGreya [7K]

ans is: 1s2, 2s2 ,2p1,...

7 0
3 years ago
8. Estimate the maximum volume percent of Methanol vapor that can exist at standard conditions. Vapor pressure = 88.5 mm Hg in a
sashaice [31]

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

It is known that at standard condition, vapor pressure is 760 mm Hg.

And, it is given that methanol vapor pressure in air is 88.5 mm Hg.

Hence, calculate the volume percentage as follows.

                  Volume percentage = \frac{\text{given vapor pressure}}{\text{standard vapor pressure}} \times 100

                                                    = \frac{88.5}{760} \times 100

                                                    = 11.65%

Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.

6 0
3 years ago
What substances pass through a leaf's stomata?
Lera25 [3.4K]
Oxygen and carbon dioxide
7 0
3 years ago
Read 2 more answers
What is a promblem based on charles' law?
LUCKY_DIMON [66]

Answer:

You will never know the exact volume with charles law

Explanation:

Doubling the temperature of gas doubles its volume, so long as the pressure and quantity of the gas are unchanged.

5 0
4 years ago
Read 2 more answers
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