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polet [3.4K]
3 years ago
8

You are taking a 578.23 mile trip to visit your grandparents. What is the distance of this trip in kilometers?

Chemistry
1 answer:
Tamiku [17]3 years ago
5 0
578.23 miles = 930.6 kilometers (Rounded up to the nearest tenth.)
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Jenna built the electrical circuit seen here. What would happen to the electrical energy if Jenna removed one of the clips on th
otez555 [7]

Answer: Option (D) is the correct answer.

Explanation:

An electric circuit works well when all the connections are complete but if any of the connections in the circuit is loose or disconnected then it is possible that current will not flow from the circuit.

Therefore, when Jenna removed one of the clips on the battery then circuit becomes incomplete and as a result there will no flow of current.

Thus, we can conclude that the electrical energy would stop because the circuit is incomplete.

6 0
3 years ago
A chemist has two large containers of sulfuric acid solution, with different concentrations of acid in each container. Blending
vitfil [10]

Answer:

0.015% and 0.012%

Explanation:

using simultaneous equation

6 0
3 years ago
Acetaldehyde decomposes at 750 K: CH3CHO → CO + CH4. The reaction is first order in acetaldehyde and the half-life of the reacti
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Answer:

k = 1.3 x 10⁻³ s⁻¹

Explanation:

For a first order reaction the integrated rate law is

Ln [A]t/[A]₀ = - kt

where [A] are the concentrations of acetaldehyde in this case, t is the time and k is the rate constant.

We are given the half life for the concentration of acetaldehyde to fall to one half its original value, thus

Ln [A]t/[A]₀ = Ln 1/2[A]₀/[A]₀= Ln 1/2 = - kt

- 0.693 = - k(530s) ⇒ k = 1.3 x 10⁻³ s⁻¹

4 0
3 years ago
A form of electricity which can attract things
qwelly [4]

 

A form of electricity which can attract things is static electricity. Static electricity is the result of an imbalance between negative and positive charges in an object. These charges built up o the surface of an object until they find a way to e released or discharged causing the attraction of things.

8 0
3 years ago
What is the solution's freezing point: 15 g of CH4N2O (Molar mass = 60.055 g/mol) in 200. g of H2O? (Kf = 1.86 (°C·kg)/mol)
AURORKA [14]

Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :

1 mol of urea =15/60.055 = 0.25mol

therefore 200g of water contain 0.25mol

the next step is to determine the malality of our solution in 200g of water, to do this we say:

200 g = 1Kg/1000g = 0.2kg

therefor 0.25mol/0.2Kg = 1.25mol/kg

and from the equation:

we know that i = 1

we are given Kf

b is the molality that we just calculated

therefore;

the solutions freezing point is -2.325°C

8 0
1 year ago
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