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bazaltina [42]
2 years ago
8

A certain material has a mass of 241.25 g while occupying 12.5 cm3 of space. what is this material? (hint: use table 3.1).

Chemistry
1 answer:
andre [41]2 years ago
4 0
To get the density of the material, the formula would be: mass divided by volume which is denoted by d = m/v
So for this problem, the mass is 241.25 g and the volume is 12.5 cm^3
so plugging in the data, we have: d = 241.25 g / 12.5 cm^3 = 19.3 g/cm^3

The answer is Tungsten, you could see it in the table, in the attached file.

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List the number of each type of atom on the left side of the equation 2C10H22(l)+31O2(g)→20CO2(g)+22H2O(g)
ValentinkaMS [17]

Answer: There are 20 carbons atoms, 62 oxygen atoms and 44 carbon atoms on the left side of the equation.

Explanation:

In the given chemical reaction:

2C_{10}H_{22}(l)+31O_2(g)\rightarrow 20CO_2(g)+22H_2O(g)

Reactants side = Left side

Number of atoms =

Coefficient × Number of atoms of an element in a unit molecular formula

Number of carbon atoms = 2 × 10 = 20

Number of oxygen atoms = 31 × 2 = 62

Number of hydrogen atoms =  2 × 22 = 44

8 0
2 years ago
A gas balloon has a volume of 80.0 mL at 300K , and a pressure of 50.0 kPa. if the pressure changes to 80.0 kPa and the temperat
stellarik [79]

Answer: 53.3

Explanation:

V2=(T2 x P1 x V1)/(T1 x P2)

(320x50x80)/(300x80)

53.3

3 0
2 years ago
An archaeologist discovers an ancient wooden bowl. which isotope would most likely be used to determine its age?
Vesnalui [34]

Answer:

c

Explanation:

whdkcbwubcdw

7 0
2 years ago
Read 2 more answers
In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for
joja [24]

Answer:

ΔG° = -118x10³ J/mol

Explanation:

The two half-reactions in the cell are:

Oxidation half-reaction:

Co(s) → Co²⁺(aq) + 2e⁻; E° = -0,28V

Reduction half-reaction:

Cu²⁺(aq)+2e⁻ → Cu(s); E° = 0,34V

The E° of the cell is defined as:

E_{cell} = E_{red} - E_{ox}

Replacing:

0,34V - (-0,28V) = 0,62V

It is possible to obtain the keq from E°cell with Nernst equation thus:

nE°cell/0,0592 = log (keq)

Where:

E°cell is standard electrode potential (0.62 V)

n is number of electrons transferred (2 electrons, from the half-reactions)

Replacing:

0,62V×2/0,0592 = log (keq)

20,946 = log keq

keq = 8,83x10²⁰≈ 5,88x10²⁰

ΔG° is defined as:

ΔG° = -RT ln Keq

Where R is gas constant (8,314472 J/molK) and T is temperature (298K):

ΔG° = -8,314472 J/molK×298K ln5,88x10²⁰

<em>ΔG° = -118x10³ J/mol</em>

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I hope it helps!

8 0
2 years ago
The K w for water at 0 ∘ C is 0.12 × 10 − 14 M 2 . Calculate the pH of a neutral aqueous solution at 0 ∘ C . p H = Is a pH = 7.2
labwork [276]

Answer:

pH → 7.46

Explanation:

We begin with the autoionization of water. This equilibrium reaction is:

2H₂O  ⇄   H₃O⁺  +  OH⁻            Kw = 1×10⁻¹⁴         at 25°C

Kw = [H₃O⁺] . [OH⁻]

We do not consider [H₂O] in the expression for the constant.

[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷ M

Kw depends on the temperature

0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻]  → [H₃O⁺] = [OH⁻]         at 0°C

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

- log [H₃O⁺] = pH

pH = - log 3.46×10⁻⁸ → 7.46

5 0
2 years ago
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