A jogger with a mass of 60 kg is moving at 2 m/s. What is the jogger's kinetic energy?

When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 27.6 g of carbon were burned in the presence of

In-s [12.5K]

Answer:

The answer to your question is: 101.2 g of CO2

Explanation:

C = 27.6 g

O₂ = 86.5 g remained 12.9 g

O₂ that reacted = 86.5 - 12.9 = 73.6 g

C + O₂ ⇒ CO₂ The equation is balanced

27.6 73.6 ?

MW 12 32 44

Rule of three

12 g of C------------------ 44 g CO2

27.6 g C ------------------ x

x = 27.6(44)/12 = 101.2 g of CO2

32 g of O2 --------------- 44 g of CO2

73.6 g of O2 ------------ x

x = 73.6(44)/32 = 101.2 g of CO2

6 0

3 years ago

Food chains always begin with what type of organism? Why?

Inessa05 [86]

The food chain always begins with a sun without it there would be no food chain... hope this helped!

6 0

3 years ago

The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

3 years ago

Read 2 more answers

Fill in the left side of this equilibrium constant equation for the reaction of benzoic acid (HC_6H_5CO_2) with water.

il63 [147K]

Answer:

[C₆H₅COO⁻][H₃O⁺]/[C₆H₅COOH] = Ka

Explanation:

The reaction of dissociation of the benzoic acid in water is given by the following equation:

C₆H₅-COOH + H₂O ⇄ C₆H₅-COO⁻ + H₃O⁺ (1)

The dissociation constant of an acid is the measure of the strength of an acid:

HA ⇄ A⁻ + H⁺ (2)

$K_{a} = \frac{[A^{-}][H^{+}]}{[HA]}$ (3)

<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA]. </em>

So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:

$K_{a} = \frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}$

I hope it helps you!

6 0

3 years ago

Calculate the mass percent composition of carbon in C4F8

Anon25 [30]

My work to your question

4 0

4 years ago