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3 years ago

How many atoms are in 1.15 mole Cu?

Chemistry

2 answers:

xxMikexx [17]3 years ago

4 0

6.022x10^23 is Avogadro’s number. Use this whenever you work with Stoichiometry involving Atoms, formula units, or molecules. 1 mol of anything is always Avogadro’s number.

Multiply everything on the top= 6.93 x 10^23
Divide by everything on the bottom = 6.93 x 10^23
Answer: 6.93 x 10^23 atoms Cu.

fiasKO [112]3 years ago

3 0

6.93 • 10^23 atoms Cu

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Which bonds are formed when elements share electrons and form molecules?.

Nimfa-mama [501]

Answer:

A covalent bond

Explanation:

the atoms bond by sharing electrons. Covalent bonds usually occur between nonmetals. For example, in water (H2O) each hydrogen (H) and oxygen (O) share a pair of electrons to make a molecule of two hydrogen atoms single bonded to a single oxygen atom.

6 0

2 years ago

A sample of P2Cl5 contains 179 g of phosphorous, how many grams of chlorine are present? Please show step by step thx!

natima [27]

Answer:

About 512 g.

Explanation:

We are given a sample of P₂Cl₅ that contains 179 grams of phosphorus, and we want to determine the grams of chlroine that is present.

Thus, we can convert from grams of phosphorus to moles of phosphorus, moles of phosphorus to moles of chlorine, and moles of chlorine to grams of chlorine.

From the formula, there are two moles of P for every five moles of Cl. The molecular weights of P and Cl are 30.97 g/mol and 35.45 g/mol, respectively. Hence:

$\displaystyle 179\text{ g P} \cdot \frac{1\text{ mol P}}{30.97\text{ g P}} \cdot \frac{5\text{ mol Cl}}{2\text{ mol P}} \cdot \frac{35.45\text{ g Cl}}{1\text{ mol Cl}} = 512\text{ g Cl}$

In conclusion, there is about 512 grams of chlorine present in the sample.

Alternatively, we can mass percentages. The mass percent of phosphorus in P₂Cl₅ is:

$\displaystyle \% \text{P} = \frac{2(30.97)}{5(35.45) + 2(30.97)} = 25.90\%$

Because there are 179 grams of phosphorus, the total amount of sample present is:

$\displaystyle \begin{aligned} 25.90\% \cdot m_T & = 179\text{ g P} \\ \\ m_T & = 691.1 \text{ g}\end{aligned}$

Therefore, the amount of chlorine present is 691.1 g - 179 g, or about 512 g, in agreement with our above answer.

8 0

2 years ago

A 0.1014 g sample of a purified compound containing C, H, and, O was burned in a combustion apparatus and produced 0.1486 g CO2

Alina [70]

Answer: The empirical formula for the given compound is $CH_2O$

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 $CO_2=0.1486g$

Mass of $H_2O=0.0609g$

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 0.1486 g of carbon dioxide, $\frac{12}{44}\times 0.1486=0.0405g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.0609 g of water, $\frac{2}{18}\times 0.0609=0.00677$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1014) - (0.0405 + 0.00677) = 0.054 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.0405g}{12g/mole}=0.003375moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.00677g}{1g/mole}=0.00677moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.054g}{16g/mole}=0.003375moles$

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 0.485 moles.

For Carbon = $\frac{0.003375}{0.003375}=1$

For Hydrogen = $\frac{0.00677}{0.003375}=2.00\times 2$

For Oxygen = $\frac{0.003375}{0.003375}=1$

Step 3: Taking the mole ratio as their subscripts.

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

The empirical formula for the given compound is $C_1H_2O_1=CH_2O$

Thus, the empirical formula for the given compound is $CH_2O$

8 0

3 years ago

NEED HELP ASAP!!! Please choose correctly!! Will mark brainliest thank you!!!

almond37 [142]

You will have a excess gas

4 0

3 years ago

When a distance between gas particles increases, the volume of the gas increases true or false??

Studentka2010 [4]

It is true that when a distance between gas particles increases, the volume of the gas will also increase.

Answer: Option A

Explanation:

The volume of any material can be the space occupied by any material. And the space required by any material to occupy a desired shape or size is determined by its bonding and distance between the particles making that material. If the particles which are soft and small can be compressed easily to occupy very small volume compared to compound made up of hard and large particles.

Along with this, the distance between the neighbouring particles also play a major role, as if we consider a laddo, we can compress it to small size by reducing the distance between neighbouring particles while we will not be able to compress the size or volume occupied by a table as we cannot reduce the distance of particle separation. Thus, it is true that if the distance between gas particles increased then the volume of gas particles will also increase.

5 0

3 years ago