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ZanzabumX [31]
3 years ago
11

A is 3600 m away from B. Jack spent 720.0 s running from A to

Chemistry
1 answer:
laiz [17]3 years ago
3 0

Answer:

Jacks average speed is 5 s per 1 m

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Use the information from the diagram to calculate the enthalpy of combustion for methane. (HELP ASAPPP!!)
Blizzard [7]

Answer:

where is the diagram sisters

6 0
3 years ago
Which of the following is NOT true of white dwarfs?
liq [111]

Answer:

A white dwarf is what stars like the Sun become after they have exhausted their nuclear fuel. Near the end of its nuclear burning stage, this type of star expels most of its outer material, creating a planetary nebula. Only the hot core of the star remains. ... That means a white dwarf is 200,000 times as dense.

Explanation:

they are cold

they are about the size of Earth

these both are not true

6 0
3 years ago
Read 2 more answers
Which element has the lowest first ionization energy? (F), (Li), (Fr), (Rb)?
Mariulka [41]

Answer:

Fr

Explanation:

Francium (Fr) has the lowest ionization energy, with a value of 4.0727eV. This ionization energy increases as one advances in a period, being greater in the group of nonmetals.

6 0
3 years ago
Question 2: you have two systems: 1) solid gold and water 2) solid sodium and waterWhich one can spontaneously release energy an
Valentin [98]

Hey there!

Solid Sodium and water will react spontaneously and release energy.  This is based on the reactivity series. Sodium is a highly reactive metal and hence, it is placed at the top of the reactivity series. This is because it loses its outermost electron very readily. When it comes in contact with water, it reacts with it violently to form sodium hydroxide and hydrogen gas. This reaction is exothermic and hence, accompanied with a release of energy.  Gold lies at the bottom of the reactivity series as it is very stable and does not give away its outermost electrons easily. Therefore, when it comes in contact with water, there is no reaction and no release of energy.

7 0
4 years ago
2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0
3 years ago
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