Why does the graph through the point (0,0)

trial 2, 2.68 g/cm/3 trial 3, 2.84g/cm/3. aluminum has a density of 2.70g/cm/3 calcute the percent error for each trial.

serg [7]

Answer:

trial 2: 0.74

trial 3: 5.19

I think but the equation for sloving percent error is:

(true value - determined value)/true value * 100

Explanation:

6 0

3 years ago

Why are some elements no longer found on earth

Iteru [2.4K]

Answer:

A synthetic element is one of 24 chemical elements that do not occur naturally on Earth: they have been created by human manipulation of fundamental particles in a nuclear reactor, a particle accelerator, or the explosion of an atomic bomb; thus, they are called "synthetic", "artificial", or "man-made".

Explanation: hope that helps

3 0

3 years ago

Read 2 more answers

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

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 $CO_2=3.095g$

Mass of $H_2O=1.902g$

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 $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

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 $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago

What volume would 0.853 moles of Nitrogen gas occupy at STP?

beks73 [17]

Answer:

19.12 L

Explanation:

At STP(i.e. Standard temperature and pressure).

The volume occupied by one mole of gas = 22.4 L

The pressure = 1 atm

The temperature = 273 K

Thus, since 1 mole of gas = 22.4 L;

Then 0.853 moles of N2 gas will occupy:

= (0.853 moles of N2 gas × 22.4 L)/ 1 mole of N2 gas

= 19.12 L

6 0

3 years ago

Accounting Concepts Should we buy stock in a company? Should we extend a line of credit to a company? Should we continue with cu

CaHeK987 [17]

Answer:

1

Explanation:

3 0

3 years ago