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2 years ago

5

Have a definite volume, but move quickly enough to overcome the forces of attraction

2 answers:

tiny-mole [99]2 years ago

8 0

Answer:

Liquid

Explanation:

I just got it right (:

patriot [66]2 years ago

6 0

Answer:

do you have a pic

Explanation:

You might be interested in

"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"

Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical</em>

<em />

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

<em>pKa NH₃/NH₄⁺</em>

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = <em>4.74</em>

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

<em>Moles NH₃</em>

<em>2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃</em>

<em>H-H equation:</em>

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 = log [0.400 moles] / [NH₄Cl]

0.0087 = [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

<em />

3 0

3 years ago

3.2 moles of H3PO4 to grams

Nimfa-mama [501]

Answer:

313, 6grams of H3PO4

Explanation:

We calculate the weight of 1 mol of H3PO4:

Weight 1 mol H3PO4= (Weight H)x3+ (Weight P)+(Weight 0)x4 =1gx3+31g+16gx4

Weight 1 mol H3PO4=98 g /mol

1 mol-----98 grams H3PO4

3,2mol----x= (3,2molx 98 grams H3PO4)/ 1mol=313,6 grams H3PO4

4 0

2 years ago

Farmers who raise cotton once used arsenic acid, H₃AsO₄, as a defoliant at harvest time. Arsenic acid is a polyprotic acid with

Stels [109]

Explanation:

The reaction equation will be as follows.

$H_{3}AsO_{4} \rightleftharpoons H_{2}AsO^{-}_{4} + H^{+}$

Hence, the expression for $K_{a}$ is as follows.

$K_{a} = \frac{[H_{2}SO^{-}_{4}][H^{+}]}{[H_{3}AsO_{4}]}$

Let us assume that the concentration of both $[H_{2}AsO^{-}_{4}]$ and $[H^{+}]$ is x.

$2.5 \times 10^{-4} = \frac{x \times x}{0.5}$

x = 0.01118034

This means that the concentration of $[H^{+}]$ is 0.01118034.

Since, we know that the relation between pH and concentration of hydrogen ions is as follows.

pH = $-log [H^{+}]$

= $-log (0.01118034)$

= 1.958

Thus, we can conclude that the pH of a 0.500 M solution of arsenic acid is 1.958.

6 0

3 years ago

2. What human-related environmental

WITCHER [35]

Answer:

Polution

Explanation:

7 0

2 years ago

107.854 how many significant

sattari [20]

Answer:

3

Explanation:

look after DP theres 3 digits

6 0

3 years ago