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brilliants [131]
4 years ago
5

Ammonia (nh3) involves an unequal sharing of electrons between nitrogen and three hydrogen atoms. what type of bonding does ammo

nia have?
Chemistry
1 answer:
Crank4 years ago
7 0
The molecules in ammonia are bonded together by covalent bond; this  is the type of bond in which electrons are shared among atoms in the compound. Ammonia is a polar molecule and is made up of one nitrogen atom and three hydrogen atoms. The hydrogen atom in ammonia is bonded to the three hydrogen atoms via sharing of three electron pairs, one with each atom of hydrogen. For each of the three covalent bond formed, one electron is supplied by nitrogen and one by hydrogen atom. 
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Which of the following tables correctly identifies each change as chemical or physical?
Anna [14]

Answer:

A is the answer

Explanation:

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3 years ago
What is the answer to this? i do give brainliest to the first person who responds!!!!!
IrinaVladis [17]

Answer: the moon passes through the earth's shadow

Explanation:

In this case the earth's shadow blocks the sun's light which normally is reflected off the moon's surface. So since the sunlight is blocked by the earth we call this a lunar eclipse

4 0
3 years ago
Two rocks one weighing 100m and the other weiging 200n are drpped from a 50-m cliff at the same time when both rocks are 10m fro
Ratling [72]

Answer:

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3 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibrium described by the equation N2O4(g)
lora16 [44]

<u>Answer:</u> The value of equilibrium constant is 0.997

<u>Explanation:</u>

We are given:

Percent degree of dissociation = 39 %

Degree of dissociation, \alpha = 0.39

Concentration of N_2O_4, c = \frac{1mol}{1L}=1M

The given chemical equation follows:

                     N_2O_4\rightleftharpoons 2NO_2

<u>Initial:</u>                c             -

<u>At Eqllm:</u>         c-c\alpha      2c\alpha

So, equilibrium concentration of N_2O_4=c-c\alpha =[1-(1\times 0.39)]=0.61M

Equilibrium concentration of NO_2=2c\alpha =[2\times 1\times 0.39]=0.78M

The expression of K_{c} for above equation follows:

K_{c}=\frac{[NO_2]^2}{[N_2O_4]}

Putting values in above equation, we get:

K_{c}=\frac{(0.78)^2}{0.61}\\\\K_{c}=0.997

Hence, the value of equilibrium constant is 0.997

4 0
3 years ago
Calculate the mass % of K in potassium phosphate
Blizzard [7]

Answer:

\large \boxed{\mathbf{1. \, \,55.26 \, \%; \, \,2. \, \,30.15 \, \%  }}

Explanation:

1. Mass percent of K

\begin{array}{cccr}\mathbf{Atoms} & \textbf{Contribution} &   & \textbf{Subtotal}\\\text{3K} & 3 \times 39.10 & = & 117.30\\\text{1P} & 1 \times 30.97 & = & 30.97\\\text{4O} & 4 \times 16.00 & = &64.00\\& \text{TOTAL} & = & \mathbf{212.27}\\\end{array}

\text{Percent K} =  \dfrac{\text{117.30 g}}{\text{212.27 g}} \times \, 100\% =  \mathbf{55.26 \, \%}\\\text{The percent by mass of potassium is $\large \boxed{\mathbf{55.26 \, \% }}$}

2. Mass percent of O

\text{Percent O} =  \dfrac{\text{64.00 g}}{\text{212.27 g}} \times \, 100\% =  \mathbf{30.15 \, \%}\\\text{The percent by mass of oxygen is $\large \boxed{\mathbf{30.15 \, \% }}$}

6 0
3 years ago
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