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Lorico [155]
3 years ago
12

How was the periodic table arranged

Chemistry
2 answers:
AysviL [449]3 years ago
7 0
By atomic number. In case you need to know it was arranged by a man named Mendeleev
Andrews [41]3 years ago
5 0
The elements on the periodic table are arranged in order, by increasing atomic number.<span />
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Another metal phosphate is cobalt phosphate. It will behave similar to calcium phosphate in an acid solution. What is the net io
ryzh [129]
We are given with
Cobalt phosphate - CoPO4

We are asked for the net ionic equation for the phosphate dissolving in H3O+

The net ionic equation is
CoPO4 (s) + H3O+ (aq) ----->  HPO42- (aq) + Co3+ (aq) + H2O *(l)
8 0
3 years ago
Read 2 more answers
A decorative "ice" sculpture is carved from dry ice (solid CO2) and held at its sublimation point of –78.5°C. Consider the proce
Leno4ka [110]

Answer:

The answers to the questions are;

a. The entropy of sublimation for carbon dioxide (the system) is  

134.07 J/Kmol.

b. The entropy of the universe for this reversible process is 376 J/K.

Explanation:

Entropy of sublimation is the entropy change experienced following the transformation of a mole of solid to vapor at  the temperature where the sublimation is taking place

a. We note that the mass of the solid CO₂ = 389 g

Molar mass of CO₂ = 44.01 g/mol

Number of moles of CO₂ in the sculpture = Mass/(Molar mass)

= (389 g)/(44.01 g/mol) = 8.84 Moles

Entropy of sublimation is given by

ΔS_{sublimation} = S_{vapor} - S_{solid} = \frac{\Delta H_{sublimation}}{T}

Where:

ΔH_{sublimation}  = 26.1 KJ/mol

T = Temperature = –78.5°C = ‪194.65‬ K

Therefore the amount of heat required to cause the 389 g of dry ice to sublime =    26.1 KJ/mol  × 8.84 Moles = 230.695 KJ

Therefore the entropy of sublimation = ΔS_{sublimation} = \frac{230.695 KJ}{194.65 K}

= 1.185 KJ/K

= 1185 J/K = 1185/8.84 J/Kmol = 134.07 J/Kmol

b. The entropy of the universe is given by;

ΔS_{universe} = \Delta S_{system} + ΔS_{surrounding}  

If the heat absorbed by the system is the same as the heat given off by the surrounding, then we have;

ΔS_{universe} = \frac{Q}{ T_{system}}  -\frac{Q}{T_{surrounding}}  

                =1.185 KJ/K - -\frac{230.695 KJ}{285.15K} = 1.185 KJ/K - 0.809 KJ/K = 0.376 KJ/K

= 376 J/K.

7 0
3 years ago
Calculate the vapor pressure of water above a solution prepared by dissolving 28.5 g of glycerin (c3h8o3) in 135 g of water at 3
pshichka [43]
Data:

<span>Solute: 28.5 g of glycerin (C3H8O3)
Solvent: 135 g of water at 343 k.
Vapor pressure of water at 343 k: 233.7 torr.

Quesiton: Vapor pressure of water

Solution:

Raoult's Law: </span><span><span>The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature  multiplied by its mole fraction.

Formula: p = Xsolvent * P pure solvent

X solvent = moles solvent / moles of solution

molar mass of H2O = 2*1.0g/mol + 16.0 g/mol = 18.0 g/mol

moles of solvent = 135 g of water / 18.0 g/mol = 7.50 mol

molar mass of C3H8O3 = 3*12.0 g/mol + 8*1 g/mol + 3*16g/mol = 92 g/mol

moles of solute = 28.5 g / 92.0 g/mol = 0.310 mol

moles of solution = moles of solute + moles of solvent = 7.50mol + 0.310mol = 7.810 mol

Xsolvent = 7.50mol / 7.81mol = 0.960

p = 233.7 torr * 0.960 = 224.4 torr

Answer: 224.4 torr
</span> </span>
8 0
4 years ago
11. Calculate the number of atoms in 2.00g of platinum.
Serggg [28]

Answer:

Explanation:

All you need to know is the atomic mass of platinum, and Avogadro's number.

2.00g Pt divided by atomic mass gives you the moles of platinum, and multiplying by avogadro's number (6.022 x 10^23) gives you the number of atoms.

Atomic mass of platinum can be found on any periodic table.

Hope this helped.

7 0
4 years ago
Read 2 more answers
Find the weighted average of these values. Col1 Value 5.00 6.00 7.00Col2 Weight 75.0 % 15.0 % 10.0 %
Nadya [2.5K]

Answer:

5.35

Explanation:

Value   5.00        6.00       7.00

Weight 75.0%     15.0%      10.0 %

We can determine the weighted average of these values using the following expression.

Weighted average = ∑ wi × xi

where,

w: relative weight

x: value

Weighted average = 5.00 × 0.750 + 6.00 × 0.150 + 7.00 × 0.100

Weighted average = 5.35

7 0
3 years ago
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