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vladimir2022 [97]
1 year ago
6

Suppose that 1500 kJ of energy were transferred to water at 20.0°C. What mass of water could be brought to the boiling point? He

at capacity (c) for liquid water is 4.18 J/g C​
Chemistry
1 answer:
FromTheMoon [43]1 year ago
8 0

That 1500 kJ of energy were transferred to water at 20.0°C. Heat capacity (c) for liquid water is 4.18 J/g C​. mass of water could be brought to the boiling point is 4485 g.

given that :

heat energy = 1500 kJ

heat capacity , c = 4.18 J/g °C

initial temperature = 20.0°C

boiling of water ,final temperature = 100 °C

Q = mcΔT

m = Q / (cΔT)

m = 1500 / ( 4.18 × ( 100 °C - 20 °C )

m = 1500 / 334.4

m = 4.485 kg = 4485 g

Thus, That 1500 kJ of energy were transferred to water at 20.0°C. Heat capacity (c) for liquid water is 4.18 J/g C. ​mass of water could be brought to the boiling point is 4485 g.

To learn more about heat capacity here

brainly.com/question/17058254

#SPJ1

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Using the periodic tables, identify the heaviest member of each of the following groups:
Lady_Fox [76]

Answer:

(a) Alkali metals: Francium (Fr)

(b) Chalcogens: Polonium (Po)

(c) Noble gases: Radon (Rn)

(d) Alkaline earth metals: Radium (Ra)

Explanation:

In the periodic table, the atomic mass increases down the group. Therefore, the last element of a group is the heaviest element of the group.

(a) alkali metals: The chemical elements that are present in group 1 of the periodic table, except hydrogen.

<u>The heaviest member of this group is francium (Fr)</u>

(b) chalcogens: The chemical elements that are present in group 16 of the periodic table

<u>The heaviest member of this group is polonium (Po)</u>

(c) noble gases:  The chemical elements that are present in group 18 of the periodic table

<u>The heaviest member of this group is radon (Rn)</u>

(d) alkaline earth metals: The chemical elements that are present in group 2 of the periodic table.

<u>The heaviest member of this group is radium (Ra)</u>

7 0
3 years ago
The molar heat of fusion for water is 6.01 kJ/mol. How much energy must be added to a 75.0-g block of ice at 0°C to change it to
andreyandreev [35.5K]
Answer is: 25,06 kJ of energy must be added to a 75 g block of ice.
ΔHfusion(H₂O) = 6,01 kJ/mol.
T(H₂O) = 0°C.
m(H₂O) = 75 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 75 g ÷ 18 g/mol.
n(H₂O) = 4,17 mol.
Q = ΔHfusion(H₂O) · n(H₂O)
Q = 6,01 kJ/mol · 4,17 mol
Q = 25,06 kJ.
7 0
3 years ago
Read 2 more answers
A pure gold ring weighs 23.5 grams. How many atoms of gold<br> are in the ring?
Digiron [165]

The number of atoms of gold in the pure ring are 7.18 × 10²² atoms.

<h3>HOW TO CALCULATE NUMBER OF ATOMS?</h3>

The number of atoms in a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

The number of moles in the gold (Au) can be calculated by dividing the mass of gold by its molar mass (196.97g/mol).

no. of moles = 23.5g ÷ 196.97g/mol

no. of moles = 0.119mol

Number of atoms in Au = 0.119 × 6.02 × 10²³

no. of atoms = 7.18 × 10²² atoms.

Therefore, the number of atoms of gold in the pure ring are 7.18 × 10²² atoms.

Learn more about number of atoms at: brainly.com/question/15959704

3 0
3 years ago
The density of thorium, which has the FCC structure, is 11.72 g/cm3. The atomic weight of thorium is 232 g/mol. Calculate (a) th
seropon [69]

Answer:

(a) a = 5.08x10⁻⁸ cm

(b) r = 179.6 pm  

Explanation:

(a) The lattice parameter "a" can be calculated using the following equation:

\rho = \frac{(N atoms/cell)*m}{V_{c}*N_{A}}      

<em>where ρ: is the density of Th = 11.72 g/cm³, N° atoms/cell = 4, m: is the atomic weight of Th = 232 g/mol, Vc: is the unit cell volume = a³, and </em>N_{A}<em>: is the Avogadro constant = 6.023x10²³ atoms/mol. </em>

Hence the lattice parameter is:  

a^{3} = \frac{(N atoms/cell)*m}{\rho *N_{A}} = \frac{4 atoms*232 g/mol}{11.72 g/cm^{3} *6.023 \cdot 10^{23} atoms/mol} = 1.32 \cdot 10^{-22} cm^{3}

a = \sqrt[3]{1.32 \cdot 10^{-22} cm^{3}} = 5.08 \cdot 10^{-8} cm

(b) We know that the lattice parameter of a FCC structure is:

a = \frac{4r}{\sqrt{2}}

<em>where r: is the atomic radius of Th</em>

Hence, the atomic radius of Th is:

r = \frac{a*\sqrt{2}}{4} = \frac{5.08 \cdot 10^{-8} cm*\sqrt{2}}{4} = 1.796 \cdot 10^{-8} cm = 179.6 pm    

I hope it helps you!    

4 0
3 years ago
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How much Magnesium is contained in 35.0g of Magnesium Chloride?
Yanka [14]

Answer:

There are

2.52

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10

23

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7 0
3 years ago
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