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Simora [160]
3 years ago
11

41Ca decays by electron capture. The product of this reaction undergoes alpha decay. What is the product of this second decay re

action? 41Ca decays by electron capture. The product of this reaction undergoes alpha decay. What is the product of this second decay reaction? Ti Cl Sc Ar Ca
Chemistry
1 answer:
dimaraw [331]3 years ago
4 0

Answer:

Cl

Explanation:

⁴¹Ca has an atomic number equal to 20, it means that it has 20 protons and 20 electrons ad its neutral state. In the decay by electron capture, it will lose one electron and will become a cation, but the mass (41) and the atomic number will remain the same.

When Ca⁺ undergoes alpha decay, it will lose an alpha particle, which has mass 4 and 2 protons.

⁴¹₂₀Ca⁺ → ³⁷₁₈X⁺ + ⁴₂α

To be stable, X will lose a proton and will become ³⁷₁₇X. The element which has atomic number 17 is chlorine, Cl.

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Determine the mass of water in kg, if 4300 cal of energy is placed in water, resulting in a temperature change to 101.0 oC from
Airida [17]

Answer:

0.5059kg

Explanation:

The heat absorbed for the water is determined using the equation:7

Q = C×m×ΔT

<em>Where Q is heat absorbed (4300cal)</em>

<em>C is specific heat (1cal/g°C)</em>

<em>m is the mass in grams</em>

<em>ΔT is change in °C (101.0°C - 92.5°C = 8.5°C)</em>

<em />

Replacing:

4300cal = 1cal/g°C×m×8.5°C

505.9g = m

In kg, the mass of water is:

<h3>0.5059kg</h3>

<em />

6 0
3 years ago
A 5g piece of aluminum foil (c=0.897) at 100°C is dropped into a 25g container of water (c=4.184) at 20°C. What is the final tem
jeyben [28]

Answer:

23.6°C

Explanation:

use delta h method

5 0
3 years ago
what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c
slava [35]

Answer : The specific heat of the substance is 0.0936 J/g °C

Explanation :

The amount of heat Q can be calculated using following formula.

Q = m \times C \times \bigtriangleup T

Where Q is the amount of heat required = 300 J

m is the mass of the substance = 267 g

ΔT is the change in temperature = 12°C

C is the specific heat of the substance.

We want to solve for C, so the equation for Q is modified as follows.

C = \frac{Q}{m \times \bigtriangleup T}

Let us plug in the values in above equation.

C = \frac{300J}{267g \times 12 C}

C = \frac{300J}{3204 g C}

C = 0.0936 J/g °C

The specific heat of the substance is 0.0936 J/g°C

3 0
3 years ago
Who are the scientists that contributed in arranging of the periodic table?
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7 0
2 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
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