The current periodic table arranges elements by

A 40.2 g sample of a metal heated to 99.3Â°C is placed into a calorimeter containing 120 g of water at 21.8Â°C. The final temper

aliina [53]

Answer:

B) Iron (c=0.45 J/g°C)

Explanation:

Given that:-

Heat gain by water = Heat lost by metal

Thus,

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

Where, negative sign signifies heat loss

Or,

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

For water:

Mass = 120 g

Initial temperature = 21.8 °C

Final temperature = 24.5 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 40.2 g

Initial temperature = 99.3 °C

Final temperature = 24.5 °C

Specific heat of metal = ?

So,

$120\times 4.184\times (24.5-21.8)=40.2\times C_{metal}\times (99.3-24.5)$

$40.2C_{metal}\left(99.3-24.5\right)=120\times \:2.7\times \:4.184$

$40.2C_{metal}\left(99.3-24.5\right)=1355.616$

$C_{metal}=0.45\ J/g^0C$

<u>This value corresponds to iron. Thus answer is B.</u>

3 0

3 years ago

How is ti possible that, when adding protons while moving left to right across a period, the size of an atom shrinks? Why is the

Kaylis [27]

As one moves across a period, from left to right, both the number of protons and electrons of a neutral atom increase. The enhancing number of electrons and protons results in a greater attraction between the electrons and the nucleus. This uplifted attraction pulls the electrons nearer to the nucleus, therefore, reducing the size of the atom.

On the other hand, while moving down a group, there is an increase in the number of energy levels. The enhanced number of energy levels increases the size of the atom in spite of the elevation in the number of protons. In the outermost energy levels, the protons get attracted towards the nucleus, however, the attraction is less due to an increase in the distance from the nucleus.

4 0

3 years ago

What two units are best suited to measure the dimensions of a fish tank

Sati [7]

Answer:

Centimeter and Liter :)

4 0

2 years ago

Read 2 more answers

A 1.42-g sample of a pure compound with formula m2so4 was dissolved in water and treated with an ex- cess of aqueous calcium chl

solong [7]

The reaction between $m_{2}SO_{4}$ with calcium chloride can be shown as- $m_{2}SO_{4}$ + $CaCl_{2}$ → $CaSO_{4}$ ↓+2mCl. The molecular weight of $CaSO_{4}$ is 136.14g. The weight of sulfate ion is 96.06g. The molecular weight of $m_{2}SO_{4}$ = (2×m + 96.06). From the reaction we can see that 1 mole of calcium chloride reacts with 1 moles $m_{2}SO_{4}$ to produce 1 mole of calcium sulfate. Now 1.36g of calcium sulfate is equivalent to 1.36/136.14=9.989× $10^{-3}$ moles of calcium sulfate.