This formula equation is unbalanced

The number of waves that pass a point in one second is called ________________________________

almond37 [142]

The characteristics of a wave are given certain names to describe them, and this helps scientists to accurately describe any given wave. The characteristic known as frequency describes the number of waves that pass a point, and it is measured in waves per second, or Hertz, which is given the symbol Hz, but can also be described using the inverse of the SI unit for second, s^-1.

3 0

3 years ago

List the major types of intermolecular forces in order of increasing strength. Is there some overlap? That is, can the strongest

Doss [256]

Answer:

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Explanation:

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6 0

3 years ago

Which term describes the composition of matter shown in figure on the left? element compound mixture phase

dangina [55]

Answer:

bb

Explanation:

6 0

2 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

3 years ago

PLEASE HURRY AND SHOW WORK

IgorC [24]

Answer:

The answer to your question is P = 1.64 atm

Explanation:

Data

Volume = 2.5 x 10⁷ L

Temperature = 22°C

Pressure = ?

Moles = 1.7 x 10⁶

R = 0.082 atm L/ mol°K

Process

1.- Convert temperature to °K

Temperature = 22 + 273

= 295°K

2.- Use the Ideal gas law to solve this problem

PV = nRT

- Solve for P

P = nRT / V

- Substitution

P = (1.7 x 10⁶)(0.082)(295) / 2.5 x 10⁷

- Simplification

P = 41123000 / 2.5 x 10⁷

- Result

P = 1.64 atm

3 0

3 years ago