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FinnZ [79.3K]
3 years ago
12

In rows by, elements with the same number of

Chemistry
1 answer:
Simora [160]3 years ago
3 0

Answer:

All elements in a row have the same number of electron shells. Each next element in a period has one more proton and is less metallic than its predecessor. Arranged this way, groups of elements in the same column have similar chemical and physical properties, reflecting the periodic law.

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Calculate either [H3O+] or [OH−] for each of the solutions at 25 °C. Solution A: [OH−]=1.55×10−7 M Solution A: [H3O+]= M Solutio
Ahat [919]

Answer: A. [H_3O^+]=0.64\times 10^{-7}M

B. [OH^-]=0.11\times 10^{-5}M

C. [OH^-]=\frac{10^{-14}}{0.000775}=1.3\times 10^{-11}M

Thus solution B is basic in nature.

Explanation:

pH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

pH for acidic solutions is less than 7, for basic solutions it is more than 7 and for neutral solutions it is equal to 7.

pH=-\log [H^+]

pOH=-log[OH^-]

pH+pOH=14

or [H^+][OH^-]=10^{-14}

A. [OH^-]=1.55\times 10^{-7}M

[H_3O^+]=\frac{10^{-14}}{1.55\times 10^{-7}}=0.64\times 10^{-7}M

pH=-log[H_3O^+]=-log[0.64\times 10^{-7}]=7

B. [H_3O^+]=9.43\times 10^{-9}M

[OH^-]=\frac{10^{-14}}{9.43\times 10^{-9}}=0.11\times 10^{-5}M

pH=-log[H_3O^+]=-log[9.43\times 10^{-9}]=8

C. [H_3O^+]=0.000775M

[OH^-]=\frac{10^{-14}}{0.000775}=1.3\times 10^{-11}M

pH=-log[H_3O^+]=-log[0.000775]=3

Thus solution B is basic.

4 0
4 years ago
What mass of silver chloride (m = 143.4) will dissolve in 1.00 l of water? the ksp of agcl is 1.8 × 10–10 ?
trapecia [35]
Given that solubility product of AgCl = 1.8 X 10^-10

Dissociation of AgCl can be represented as follows,

AgCl(s)             ↔      Ag+(ag)             +           Cl-(aq)

Let, [Ag+] = [Cl-] = S

∴Ksp = [Ag+][Cl-] = S^2

∴ S = √Ksp = √(1.8 X 10^-10) = 1.34 x 10^-5 mol/dm3

Now, Molarity of solution = \frac{\text{weight of solute}}{\text{Molecular weight X volume of solution (l)}}
              ∴  1.34 x 10^-5     = \frac{\text{weight of AgCl}}{143.4 X 1}
∴ Weight of AgCl present in solution = 1.92 X 10^-3 g

Thus, mass of AgCl that will dissolve in 1l water = 1.92 x 10^-3 g
7 0
3 years ago
Rosa was looking for patterns to help predict the products of chemical reactions. She recorded three similar decomposition react
rewona [7]

Answer:

2LiCl + 3O2  

Explanation:

8 0
3 years ago
Read 2 more answers
What is the % of water in (MgSO4 . 2H2O)?
Grace [21]

Answer: A hydrate is found to have the following percent composition: 48.8% MgSO4 and 51.2% water.

Explanation:

i dont know

7 0
3 years ago
2.122 g of a solid mixture containing only potassium carbonate (FW = 138.2058 g/mol) and potassium bicarbonate (FW = 100.1154 g/
aleksandr82 [10.1K]

Answer:

The weight percent of potassium carbonate is 50,8 wt% and of potassium bicarbonate 49,2 wt%

Explanation:

The reactions of potassium carbonate (K₂CO₃) and potassium bicarbonate (KHCO₃) with HCl produce:

K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O

KHCO₃ + HCl → KCl + CO₂ + H₂O

That means that you need 2 moles of HCl to titrate potassium carbonate and 1 mol to titrate potassium bicarbonate.

The moles of HCl to titrate the mixture are:

0,03416L×\frac{0,762mol}{1L} = <em>0,02603 mol of HCl</em>

If X is mass of K₂CO₃ and Y is mass of KHCO₃ in the mixture, the moles of HCl to titrate the mixture are equals to:

0,02603 mol = 2X×\frac{138,2058 g}{1mol} + Y×\frac{100,1154 g}{1mol} <em>(1)</em>

As the mass of the mixture is 2,122g:

2,122g = X + Y <em>(2)</em>

Replacing (2) in (1):

0,02603 mol = 0,01447 (2,122-Y) + 9,988x10⁻³Y

0,02603 mol = 0,0307 - 0,01447Y + 9,988x10⁻³Y

-4,6778x10⁻³ = -4,4827x10⁻³Y

1,044g = Y <em>-mass of potassium bicarbonate-</em>

Thus:

X = 1,078g <em>-mass of potassium carbonate-</em>

The weight percent of potassium carbonate is:

\frac{1,078g}{2,122g}×100 =<em> 50,8 wt%</em>

The weight percent of potassium bicarbonate is:

\frac{1,044g}{2,122g}×100 = <em>49,2 wt%</em>

<em></em>

I hope it helps!

5 0
3 years ago
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