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3 years ago

14

True or false: Observing a physical property of a substance will change the substances' identity I NEEED HELLLPPPPP NOOOWWWW !!!

! PLEASEEEEE

1 answer:

andrew11 [14]3 years ago

8 0

Answer:

false

Explanation:

a physical property can be observed and stay the same but a chemical property can't

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Which group of factors are most important in determining the composition of ocean water? a.)temperature, salinity, and density b

Elis [28]

Temperature, salinity, and density are the group of factors are most important in determining the composition of ocean water.

a.)temperature, salinity, and density

<u>Explanation:</u>

The three fundamental factors that help in determining the composition of ocean water are temperature, salinity, and density. Temperature, saltiness, salinity, and density influence the thickness of seawater.

Enormous water masses of various densities are significant in the layering of the sea water (increasingly thick water sinks). As temperature builds water turns out to be less thick. As saltiness builds water gets denser. The temperature helps in deciding the pace of vanishing of the ocean.

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Temperature is a measure of

Triss [41]

Temperature is a measure of "Molecular movement"

In short, Your Answer would be Option B

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Troyanec [42]

Answer:

I think it is 1115 kJ but I don't see the answer

Explanation:

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3 years ago

G(1)=0 g(n) =g(n-1)+n g(2)=

Aliun [14]

G(2)=2

For this, you can plug in 2 everywhere you see an n. So the equation will read:
g(2)=g(2-1)+2 -> g(2)=g(1)+2. Since we are given g(1)=0, we can plug in 0 where we see g(1). The equation is now. g(2)=0+2. So, g(2)=2.

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3 years ago

At the start of a reaction, there are 0.0249 mol N2,

gladu [14]

Answer:

Explanation:

The reaction is given as:

$N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$

The reaction quotient is:

$Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}$

From the given information:

TO find each entity in the reaction quotient, we have:

$[NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}$

$[N_2] = \dfrac{0.024 }{3.5}$

$[N_2] = 0.006857$

$[H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}$

$[H_2] = 9.17 \times 10^{-3}$

∴

$Q_c= \dfrac{(1.834 \times 10^{-4})^2}{(0.0711)\times (9.17\times 10^{-3})^3} \\ \\ Q_c = 0.6135$

However; given that:

$K_c = 1.2$

By relating $Q_c \ \ and \ \ K_c$ , we will realize that $Q_c \ \ < \ \ K_c$

The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.

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