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Naya [18.7K]
3 years ago
14

On the periodic table, calcium and bromine are located in period 4. What is one difference between the two elements?

Chemistry
1 answer:
Dovator [93]3 years ago
4 0

Answer:

D

Explanation:

Just look at the groups and their tendencies. Or look at their electron configurations. You will notice that calcium has 2 valence electrons and bromine has 7 valence electrons.

Halogens have the most valence electrons possible without completing a shell (8 would be needed for that)

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The following situations are presented to a medical examiner. In which situation is the medical examiner most likely to gather m
kolezko [41]
B) 25-year-old male recovered from rapids after his empty kayak is sited by friends 
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8 0
3 years ago
8. (02.04 MC)
elena55 [62]

Answer:

3p^2

Explanation:

after filling 3s^2 only two electrons  left out of 14 so the next sub shell is 3p therefore ,X represents 3p^2

3 0
2 years ago
245 g water sample initially at at 32 oC absorbs 17 kcal of heat. What is the final temperature of water?
steposvetlana [31]

Answer:62.66°C or 235.66K

Explanation:Q=McpT, the energy was given in calories so you first convert to Joules by multiplying the value in calories by 4.184J.

17*4.184=71.128kJ.

71.128kJ=mcpT

71.128kJ=245*4.187*(T-Tm)

Tm is the final temperature of the mixture. The T is the temperature given which should be converted to Kelvin by adding 273...T=32+273=305K.

71128J=245*4.187*(305-Tm)

71128=312873.575-1025.815Tm

1025.815Tm=312873.575-71128

1025.815Tm=241745.58

Tm=241745.58/1025.815

Tm=235.66K

4 0
3 years ago
A horizontal pipe 15 cm in diameter and 4 m long is buried in the earth at a depth of 20 cm. The pipe-wall temperature is 75 deg
Basile [38]

Explanation:

The given data for case (1) is as follows.

  h = 20 cm = 0.2 m

Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.

              Q = kA \frac{\Delta T}{L}

  where,    A = area

                  L = length

                  k = thermal conductivity = 0.8 W/m

             \Delta T = change in temperature.

Therefore, putting the given values into the above formula as follows.

            Q = kA \frac{\Delta T}{L}

                = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{0.2}

                = 168 W

For case (2), h = 180 cm = 1.8 m

Therefore, heat lost will be calculated as follows.

                       Q = kA \frac{\Delta T}{L}

                           = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{1.8}

                           = 18.67 W

Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.

8 0
3 years ago
The concentration of hydrogen peroxide solution can be determined by
max2010maxim [7]

The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

with acidified potassium manganate(VII) solution. In this reaction the hydrogen

peroxide is oxidised to oxygen gas.

A 5.00 cm3 sample of the hydrogen peroxide solution was added to a volumetric flask

and made up to 250 cm3 of aqueous solution. A 25.0 cm3 sample of this diluted

solution was acidified and reacted completely with 24.35 cm3 of 0.0187 mol dm–3

potassium manganate(VII) solution.

Write an equation for the reaction between acidified potassium manganate(VII)

solution and hydrogen peroxide.

Use this equation and the results given to calculate a value for the concentration,

in mol dm–3, of the original hydrogen peroxide solution.

(If you have been unable to write an equation for this reaction you may assume that

3 mol of KMnO4 react with 7 mol of H2O2. This is not the correct reacting ratio.)

Answer:

2.275 M

Explanation:

The equation of the reaction is;

2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)

Let;

CA= concentration of MnO4^- =  0.0187 mol dm–3

CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of  MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since  

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution =  5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

V2 = final volume of H2O2 solution = 250 cm3

C1 = C2V2/V1

C1 = 0.0455 * 250/5

C1 = 2.275 M

8 0
3 years ago
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