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Ganezh [65]
3 years ago
13

What is the percent composition of hydrogen if a sample is found to contain 7 grams of hydrogen, 32 grams of nitrogen, and 7 gra

ms of carbon?
Chemistry
1 answer:
nata0808 [166]3 years ago
6 0

<u>Answer:</u> The percent composition of hydrogen in the sample is 15.22 %

<u>Explanation:</u>

We are given:

Mass of hydrogen = 7 grams

Mass of nitrogen = 32 grams

Mass of carbon = 7 grams

Total mass of the sample = 7 + 32 + 7 = 46 grams

To calculate the percentage composition of hydrogen in sample, we use the equation:

\%\text{ composition of hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of sample}}\times 100

Mass of sample = 46 g

Mass of hydrogen = 7 g

Putting values in above equation, we get:

\%\text{ composition of hydrogen}=\frac{7g}{46g}\times 100=15.22\%

Hence, the percent composition of hydrogen in the sample is 15.22 %

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Which of the following is a completely elastic collision? A. a ball rebounds against a wall, reversing its direction, but at onl
Dmitry_Shevchenko [17]

Answer:

Answer E.

For a collision to be completely elastic, there must be NO LOSS in kinetic energy.

We can go through each answer choice:

A. Since the ball rebounds at half the initial speed, there is a loss in kinetic energy. This is NOT an elastic collision.

B. A collision involving sticking is an example of a perfectly INELASTIC collision. This is NOT an elastic collision.

C. A reduced speed indicates that there is a loss of kinetic energy. This is NOT elastic.

D. The balls traveling at half the speed after the collision indicates a loss of kinetic energy, making this collision NOT elastic.

E. This collision indicates an exchange of velocities, characteristic of an elastic collision. We can prove this:

Let:

m = mass of each ball

v = velocity

We have the initial kinetic energy as:

KE = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2KE=21mv2+0=21mv2

And the final as:

KE = 0 + \frac{1}{2}mv^2 = \frac{1}{2}mv^2KE=0+21mv2=21mv2

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3 years ago
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Write the isotopic symbol for the following (show your work) a) An isotope of iodine whose atoms have 78 neutrons b) An isotope
morpeh [17]

<u>Answer:</u>

<u>For a:</u> The isotopic representation of iodine is _{53}^{131}\textrm{I}

<u>For b:</u> The isotopic representation of cesium is _{55}^{137}\textrm{Cs}

<u>For c:</u> The isotopic representation of strontium is _{38}^{52}\textrm{Sr}

<u>Explanation:</u>

The isotopic representation of an atom is: _Z^A\textrm{X}

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

  • <u>For a:</u>

We are given:

Number of neutrons = 78

Atomic number of iodine = 53 = Number of protons

Mass number = 53 + 78 = 131

Thus, the isotopic representation of iodine is _{53}^{131}\textrm{I}

  • <u>For b:</u>

We are given:

Number of neutrons = 82

Atomic number of cesium = 55 = Number of protons

Mass number = 55 + 82 = 137

Thus, the isotopic representation of cesium is _{55}^{137}\textrm{Cs}

  • <u>For c:</u>

We are given:

Number of neutrons = 52

Atomic number of strontium = 38 = Number of protons

Mass number = 38 + 52 = 90

Thus, the isotopic representation of strontium is _{38}^{52}\textrm{Sr}

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Does bacteria need water to live
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In terms of bonds, what would the molecule C₆H₁₂ be classified as?
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Answer:

Alkene

Explanation:

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Identify the group of elements that corresponds to each of the following generalized electron configurations and indicate the nu
Alja [10]

Answer:

(a) [Noble gas] ns² np⁵: Number of unpaired electron is 1 and belongs to group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²: Number of unpaired electron is 2 and belongs to group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹: Number of unpaired electron is 1 and belongs to group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶ : Number of unpaired electron is 6 and belongs to group 8 of the periodic table.

Explanation:

In the Periodic table, the chemical elements are arranged in 7 rows, called periods and 18 columns, called groups. They are organized in increasing order of atomic numbers.

(a) [Noble gas] ns² np⁵

As the total number of electrons in the p-orbital is 5. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 5 electrons in np orbital. <u>So there are 7 valence electrons.</u>

Therefore, this element belongs to the group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²

As the total number of electrons in the d-orbital is 2. Therefore, the number of unpaired electrons is 2.

This element has 2 electrons in ns orbital and 2 electrons in (n-1)d orbital. So there are <u>4 valence electrons.</u>

Therefore, this element belongs to the group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹

As the total number of electrons in the p-orbital is 1. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 1 electron in np orbital. So there are <u>3 valence electrons</u>.

Therefore, this element belongs to the group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶

As the total number of electrons in the f-orbital is 6. Therefore, the number of unpaired electron is 6.

This element has 2 electrons in ns orbital and 6 electrons in (n-2)f orbital. So there are <u>8 valence electrons.</u>

Therefore, this element belongs to the group 8 of the periodic table.

3 0
3 years ago
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