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kari74 [83]
3 years ago
15

How many molecules are present in this sample 0.423 mol co

Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
6 0
That’s Correct thank you for the help
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Trojans are _____.
liubo4ka [24]

Answer:

d. asteroids affected by Jupiter's gravity

Explanation:

A Trojan horse or Trojan is a type of malware that is often disguised as legitimate software. Trojans can be employed by cyber-thieves and hackers trying to gain access to users' systems. Users are typically tricked by some form of social engineering into loading and executing Trojans on their systems.

8 0
3 years ago
Read 2 more answers
Please help, any and all is very much appreciated!
iogann1982 [59]
1. True
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3 0
3 years ago
If 8.6 g of ch4 and 5.9 g of o2 react, what is the mass, in grams, of h2o that is produced?
Alenkasestr [34]
This may seem confusing because they give you two masses, but all you have to do is pick one to do the calculations. Personally, I would pick O2, since the molar mass is easier to calculate. The answer would be 3.3 g (rounded for sig figs). To get this, first take the 5.9 grams of O2 and convert it to moles by dividing by the molar mass of oxygen gas, which is 32. Then, multiply both by the mole-mole ratio, which is 2:2, or simply 1:1. After that, multiply that by 18g, which is the molar mass of water to get grams of water. 

REMEMBER, you have to write and balance the chemical equation before you can do any of that work. 
That happens to be CH4 + 2O2 => CO2 + 2H2O
7 0
3 years ago
The reaction (CH3)3CBr + OH- (CH3)3COH + Br- in a certain solvent is first order with respect to (CH3)3CBr and zero order with r
son4ous [18]

Answer and Explanation:

The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation

R = molar gas constant

K = A(e^(-Ea/RT))

Taking natural log of both sides

In K = In A - (Ea/RT)

In K = (-Ea/R)(1/T) + In A

Comparing this to the equation of a straight line; y = mx + c

y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A

a) From the question, m = (-Ea/R) = -1.10 × (10^4) K

(-Ea/R) = -1.10 × (10^4) = -11000

R = 8.314 J/K.mol

Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol

b) c = In A = 33.5

A = e^33.5 = (3.54 × (10^14))/s

c) K = A(e^(-Ea/RT))

A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol

K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s

QED!

5 0
3 years ago
What is the percentage of oxygen in Li(NO2)3
BartSMP [9]

Answer:

66.2 % of O

Explanation:

Our compound is the lithium nitrite.

LiNO₂

This salt is ionic and can be dissociated: LiNO₂ →  Li⁺ + NO₂⁻

We determine the molar mass:

molar mass of Li + 3 . molar mass of N + 6 . molar mass of O

6.94 g/mol + 3. 14 g/mol + 6 . 16 g/mol = 144.94 g/mol

The mass of oxygen contained in 1 mol of lithium nitrite is:

6 . 16 g/mol = 96 g

So the percentage of oxygen present is:

(96 g / 144.94 g) . 100 = 66.2 %

3 0
3 years ago
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