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Kruka [31]
2 years ago
8

What is the mass of 15100 mL of oxygen gas at STP?

Chemistry
1 answer:
adell [148]2 years ago
5 0

Answer:

02 has a mass of 1.78 g.

Explanation:

hope this helps

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Bronze is an alloy made of mostly copper (Cu) and tin (Sn). What type of bond forms bronze?
Snezhnost [94]

Answer:

A. Metallic bond

Explanation:

Think about it: copper and tin are both common metals. That's how we know it's a metallic bond!

Why not B: Covalent bonds are between two nonmetals.

Why not C: Ionic bonds are between a nonmetal and a metal.

Why not D: Paired bond isn't a common phrase in chemistry.

6 0
4 years ago
Why do farmers spray water over their crops before a frost?
Nezavi [6.7K]

Answer:

When water freezes and turns into ice, it releases latent heat. Then, the ice that builds up on the plant will insulate it from the colder surrounding air temperatures. Because of this, some growers choose to spray their crop with water before the freeze occurs.

Explanation:

5 0
3 years ago
Type the letter of the correct answer in the blank space. The image shows the combustion of methane. Which equation represents t
Talja [164]

Answer:

CH4 + 2O2 → CO2 + 2H2O

Explanation:

7 0
2 years ago
What are the 7 methods of separating mixtures?​
Gennadij [26K]

Methods Of Separating Mixtures

Handpicking.

Threshing.

Winnowing.

Sieving.

Evaporation.

Distillation.

Filtration or Sedimentation.

Separating Funnel.

3 0
3 years ago
Read 2 more answers
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
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