The atomic number of an element is......

How many grams are contained in 3.4 moles of potassium

BARSIC [14]

Answer:

The answer is 0.025576559594663

6 0

3 years ago

3. What is the atomic mass of phosphorous if phosphorous-29 has a percent abundance of 35.5%, phosphorous-30 has a percent abund

daser333 [38]

Answer:

The atomic mass of phosphorus is 29.864 amu.

Explanation:

Given data:

Atomic mass of phosphorus = ?

Percent abundance of P-29 = 35.5%

percent abundance of P-30 = 42.6%

Percent abundance of P-31 = 21.9%

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass / 100

Average atomic mass = (29×35.5)+(30×42.6) + (31×21.9) /100

Average atomic mass = 1029.5 + 1278 + 678.9/ 100

Average atomic mass = 2986.4 / 100

Average atomic mass = 29.864 amu.

The atomic mass of phosphorus is 29.864 amu.

5 0

3 years ago

The pressure of nitrogen gas at 35°C is changed from 0.89 atm to 4.3 atm. What will be its final temperature in Kelvin?

Alja [10]

Answer: The final temperature in Kelvin is 1488

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=0.89atm\\T_1=35^0C=(35+273)K=308K\\P_2=4.3atm\\T_2=?$

Putting values in above equation, we get:

$\frac{0.89}{308}=\frac{4.3}{T_2}\\\\T_2=1488K$

Hence, the final temperature in Kelvin is 1488

8 0

3 years ago

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 2.001 g sample of B

Alborosie

Answer: The empirical formula for the given compound is $C_{15}H_{24}O_1$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, iron 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 $CO_2=5.995g$

Mass of $H_2O=1.963g$

Mass of sample = 2.001 g

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

So, in 5.995 g of carbon dioxide, $\frac{12}{44}\times 5.995=1.635g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 1.963 g of water, $\frac{2}{18}\times 1.963=0.218g$ of hydrogen will be contained.

Mass of oxygen in the compound = (2.001) - (1.635 + 0.218) = 0.148 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.635g}{12g/mole}=0.136moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.218g}{1g/mole}=0.218moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.148g}{16g/mole}=0.0092moles$

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 0.0092 moles.

For Carbon = $\frac{0.136}{0.0092}=14.78\approx 15$

For Hydrogen = $\frac{0.218}{0.0092}=23.69\approx 24$

For Oxygen = $\frac{0.0092}{0.0092}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 15 : 24 : 1

Hence, the empirical formula for the given compound is $C_{15}H_{24}O_1$

3 0

4 years ago

What are some of the benefits of using measurements to find density? Provide 2 each. What are some of the challenges of using me

stellarik [79]

Some of the benefits of using measurements to find density include the following:

It is easy and doesn't require experiments.
The volume of both light and dense molecule can be gotten.

Some of the challenges of using measurements to find density include the following:

Inconsistency in the grind setting
Temperature changes not taken into account.

<h3>What is Density?</h3>

This is defined as the mass per unit volume of a substance or te degree of compactness of a substance and can be gotten with the use of measurements or experiments.

The measurement doesn't take the temperature changes into consideration which will lead to the result being most likely incorrect. It is also advantageous because less time is used thereby making it easier.

Read more about Density here brainly.com/question/406690

#SPJ1

7 0

2 years ago