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Harlamova29_29 [7]
3 years ago
11

Periods on the periodic table are

Chemistry
1 answer:
Roman55 [17]3 years ago
6 0

Answer:

Periods on the periodic table of elements

Explanation:

In the periodic table of elements, there are seven horizontal rows of elements called periods. The vertical columns of elements are called groups, or families. (See also The Periodic Table: Metals, Nonmetals, and Metalloids.) In each period (horizontal row), the atomic numbers increase from left to right.

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An electrode has a potential of 1.201 V with respect to a saturated silver-silver chloride electrode. What would the electrodes
fredd [130]

Answer:

The potential wrt. calomel is 1.254 V

Explanation:

Given:

Potential wrt. silver chloride E_{Ag}  = 1.201 V

Potential wrt. saturated silver chloride E = 0.197 V

Potential wrt. SCE E_{Hg} = 0.241 V

Now potential wrt. hydrogen is given by,

   = 1.201- 0.197

   = 1.004 V

And we find for potential wrt. calomel,

   = potential wrt. hydrogen + potential wrt. SEC

   = 1.004 +0.241

   = 1.254 V

Therefore, the potential wrt. calomel is 1.254 V

7 0
3 years ago
Which statement about gases is true?
Whitepunk [10]

Answer:

they are made up of hard spheres that are in random motion

7 0
3 years ago
Read 2 more answers
How did the industrial revolution contribute to global climate change
Deffense [45]

Answer:humans have put ridiculous amounts of carbon dioxide into the atmosphereThese events are linked to the mass burning of fossil fuels to meet an increase in human demand

So the answer is True

Explanation:

6 0
2 years ago
VA 19.75-g sample was heated by 12.35 calories. The specific heat of the sample is 0.125 cal/g°C. What was the initial temperatu
SOVA2 [1]

Answer:

31.9 °C  

Explanation:

The formula for the heat q absorbed by an object is

q = mCΔT where ΔT = (T₂ - T₁)

Data:

q = 12.35 cal

m = 19.75 g

C = 0.125 cal°C⁻¹g⁻¹

T₂ = 37.0 °C

Calculations

(a) Calculate ΔT

q = mCΔT

12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT

12.35 = 2.406ΔT °C⁻¹  

ΔT  = 12.35/(2.406 °C⁻¹) = 5.13 °C

(b) Calculate T₂

ΔT = T₂ - T₁

T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C

The original temperature was 31.9 °C.

 

6 0
3 years ago
Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks
charle [14.2K]

Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

4 0
3 years ago
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