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Leviafan [203]
3 years ago
8

A new stable element with an atomic number of 120 and an atomic mass of 246 is created in a particle accelerator. Enough of this

element is created to determine that it combines with chlorine in a 1:2 ratio. Element 120 should be placed in the same group as _____ on the periodic table.
Chemistry
1 answer:
deff fn [24]3 years ago
3 0

Answer:

Be,Mg,Ra etc

Explanation:

It should be palced in group 2A because as it reacts with chlorine in ratio of 1:2 . It's valancy is 2 and is metal as it react with non metal donating two electrons .

one more thing it fits there orderly

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Determine the relative amounts (in terms of volume fractions) for a 15 wt% sn-85 wt% pb alloy at 100°c. the densities of tin an
nikdorinn [45]
Volume fraction = volume of the element / volume of the alloy

Volume = density * mass

Base: 100 grams of alloy

mass of tin = 15 grams

mass of lead = 85 grams

volume = mass / density

Volume of tin = 15g / 7.29 g/cm^3 = 2.06 cm^3

Volume of lead = 85 g / 11.27 g/cm^3 = 7.54 cm^3

Volume fraction of tin = 2.06 cm^3 / (2.06 cm^3 + 7.54 cm^3) = 0.215

Volume fraction of lead = 7.54 cm^3 / (2.06 cm^3 + 7.54 cm^3) = 0.785

As you can verify the sum of the two volume fractions equals 1: 0.215 + 0.785 = 1.000
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4 years ago
A sample of food containing 27 g of fat, 48 g of carbohydrates and 20 g of protein is burned in a bomb calorimeter. In a perfect
rosijanka [135]

Answer:

38.3958 °C  

Explanation:

As,

1 gram of carbohydrates on burning gives 4 kilocalories  of energy

1 gram of protein on burning gives 4 kilocalories  of energy

1 gram of fat on burning gives 9 kilocalories of energy

Thus,

27 g of fat on burning gives 9*27 = 243 kilocalories of energy

20 g of protein on burning gives 4*20 = 80 kilocalories  of energy

48 gram of carbohydrates on burning gives 4*48 = 192 kilocalories  of energy

Total energy = 515 kilocalories

Using,

Q=m_{water}\times C_{water}\times (T_f-T_i)

Given: Volume of water = 23 L = 23×10⁻³ m³

Density=\frac{Mass}{Volume}  

Density of water= 1000 kg/m³

So, mass of the water:  

Mass\ of\ water=Density \times {Volume\ of\ water}  

Mass\ of\ water=1000 kg/m^3 \times {0.023\ m^3}  

Mass of water  = 23 kg

Initial temperature = 16°C  

Specific heat of water = 0.9998 kcal/kg°C  

515=23\times 0.9998\times (T_f-16)

Solving for final temperature as:

<u>Final temperature = 38.3958 °C  </u>

8 0
3 years ago
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