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atroni [7]
3 years ago
11

Matt's cube, after 5 trials, had an average density of 7.40 g/cm3 . His group's cube was made of

Chemistry
2 answers:
lawyer [7]3 years ago
5 0

its d zinc if your using usa test this dummy got me wrong trust me sis the answer is d

aleksandr82 [10.1K]3 years ago
4 0
The right answer for the question that is being asked and shown above is that: "B) gold." Matt's cube, after 5 trials, had an average density of 7.40 g/cm3 . His group's cube was made of <span>B) gold. </span>
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Neon has 2 isotopes. Neon-20 has a mass of 19:992 amu and
Alenkinab [10]

Answer:

20.0928.

Explanation:

The average atomic mass is (90 * 19.992 + 10* 21) / 100

= 20.0928.

6 0
3 years ago
What are some natural resources that are limited on earth
inysia [295]

Answer:

oil, ores, and fossils.

5 0
3 years ago
The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42
denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

k = Ae^{\frac{-Ea}{RT}}

<u>Where:</u>

k: is the rate constant

A: is the frequency factor    

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

4 0
3 years ago
Calculate the molality, molarity, and mole fraction of FeCl3 in a 29.5 mass % aqueous solution (d = 1.283 g/mL).
faltersainse [42]

Answer:

molality FeCl3= 2.579 molal

molarity FeCl3 = 2.333 M

mol fraction FeCl3 = 0.0444

Explanation:

Step 1: Data given

Mass % = 29.5 %

Density = 1.283 g/mL

Molar mass FeCl3 = 162.2 g/mol

Step 2: Calculate mass solution

Suppose we have 1L = 1000 mL solution

Mass solution = 1.283 g/mL * 1000 mL = 1283 grams

Step 3: Calculate mass FeCl3

Mass FeCl3 = 0.295 * 1283 grams

Mass FeCl3 =  378.485 grams

Step 4: Calculate mass of water

Mass water = 1283 - 378.485 = 904.515 grams

Step 5: Calculate moles FeCl3

Moles FeCl3 = mass FeCl3 / molar mass FeCl3

Moles FeCl3 = 378.485 grams / 162.2 g/mol

Moles FeCl3 = 2.333 moles

Step 6: Calculate moles H2O

Moles H2O = 904.515 grams / 18.02 g/mol

Moles H2O = 50.195 moles

Step 7: Mol fraction FeCl3

Mol fraction FeCl3 = 2.333 / (50.195+2.333)

Mol fraction FeCl3 =   0.0444

Step 8: Calculate molality

Molality = moles FeCl3 / mass H2O

Molality = 2.333 moles / 0.904515 kg

Molality = 2.579 molal

Step 9: Calculate molarity

Molarity = moles / volume

Molarity FeCl3 = 2.333 moles / 1 L

Molarity FeCl3 = 2.333 M

8 0
3 years ago
Phosphorus-32 has a half-life of 14.0 days. Starting with 4.00 g of 32P, how many grams will remain after 84.0 days ?
Lelu [443]
T₁=84 d
t₂=14 d
m₁=4 g

n=t₁/t₂
n=84/14=6

m₂=m₁/2ⁿ

m₂=4/(2⁶)=0.0625 g

0.0625 grams will remain after 84.0 days

3 0
3 years ago
Read 2 more answers
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