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3 years ago

Fluorine and Nitrogen react to form a covalent molecule. Which best describes the molecule they will make?

Chemistry

1 answer:

mafiozo [28]3 years ago

7 0

I think a is the correct answer

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In the Minnesota Department of Health set a health risk limit for acetone in groundwater of 60.0 μg/L . Suppose an analytical ch

siniylev [52]

Answer:

m = 4.7 μg

Explanation:

Given data:

density of acetone = 60.0 μg/L

Volume = 79.0 mL

Mass = ?

Solution:

Formula:

d = m/v

v = 79.0 mL × 1L /1000 mL

v = 0.079 L

Now we will put the values on formula:

d = m/v

60.0 μg/L = m/0.079 L

m = 60.0 μg/L × 0.079 L

m = 4.7 μg

So health risk limit for acetone = 4.7 μg

3 0

3 years ago

If a sample is thought to be several million years old, which method would best help to determine its absolute age?

Makovka662 [10]

The answer is B)Ar-40/k-40

6 0

3 years ago

Hi, I need help on the chemical names of the following two:

Lesechka [4]

Answer:

like how do you need help

3 0

3 years ago

Balance the following reaction in KOH (under basic conditions). What are the coefficients in for C3H8O2 and KMnO4 in the balance

GrogVix [38]

Answer:

Coefficient of $C_3H_8O_2=3$

Coefficient of $KMnO_4$ =8

Explanation:

We are given that a reaction in which $C_3H_8O_2$ reacts with $KMnO_4$

We have to find the coefficient of each reactants in balanced reaction

$3C_3H_8O_2(aq)+8KMnO_4(aq)\rightarrow 3C_3H_2O_4K_2(aq)+8MnO_2(aq)+2KOH+8H_2O$

Coefficient is defined the constant value multiplied with a reactant in a reaction.

Coefficient of $C_3H_8O_2$ =3

Coefficient of $KMnO_4=8$

Coefficient of $C_3H_2O_4K_2=3$

Coefficient of $MnO_2=8$

Coefficient of $H_2O=8$

Coefficient of KOH=2

Hence, Coefficient of $C_3H_8O_2=3$ and coefficient of $KMnO_4=8$

7 0

3 years ago

When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot

viktelen [127]

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the $'n^{th}'$ orbit of hydrogen atom.

Formula used :

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV$

Energy of n = 2 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

$E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV$

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

3 0

3 years ago