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3 years ago

What is the bond angle of a methane molecule that contains four hydrogen atoms bonded to one carbon atom?

2 answers:

6 0

Methane is a tetrahedral molecule where each C-H is kept as far away from the others as possible due to electron-electron repulsion, resulting in a bond angle of 109.5˚ (Technically 109˚ 28' but I would just use the decimal version if I were you).
Hope this helps!

Marta_Voda [28]3 years ago

5 0

Answer:

Explanation:on plato

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Determine the volume of fluid in the graduated cylinder shown.

alexgriva [62]

Answer:

47.3 ml

Explanation:

The graduated cylinder is shown in the image attached.

Now we have to take a good look at the cylinder, the lines between 45 and 50 are 46, 47, 48 and 49. Even though the points in between two lines weren't graduated but we can intelligently guess the correct volume by observing the upper meniscus of the liquid. Hence the answer.

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3 years ago

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If mercury (Hg) and oxygen (O2) were reacted to form mercury oxide how many molecules of each reactant and product would be need

dimulka [17.4K]

In order to get HgO you would need 2Hg+1O2=2HgO. Since oxygen is diatomic you need two when it stands alone causing you to need two mercuries to balance out the reactants and the product I hope this helps

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3 years ago

(GIVING BRAINLIEST) do what the following image says.

S_A_V [24]

Answer:

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3 years ago

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Alina [70]

C is the correct answer

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3 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:

$K_1=K_b$

$K_2=\frac{1}{K_w}$

Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

Hence, the value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

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3 years ago