What are two factors that determine an object's gravitational potential energy?

A compound is 52.0% zinc 9.6% carbon and 38.4% oxygen. Calculate the empirical formula of the compound.

skad [1K]

Answer:

So a compound is 52% Zinc(Zn), 9.6% Carbon(C), and 38.4% Oxygen (O). Let’s first start off by assuming that we have 100 g of this compound. This means that we have 52 g of Zinc, 9.6 g of Carbon, and 38.4 g of Oxygen.Zinc = 65.38 g/molCarbon = 12 g/molOxygen = 16 g/molThis means we have:52 g of Zn(1 mol Zn/65.38 g of Zn) ≈0.8 mol of Zn.9.6 g of C(1 mol C/12 g of C) = 0.8 mol of C38.4 g of O(1 mol of O/16 g of O) = 2.4 mol of O.

Explanation:

What we want to do next is divide each element by the common factor of all of them, which is 0.8. In most cases, you divide each element by the element with the least amount of moles. After we divide each by 0.8, you’ll notice you have 1 Zn, 1 C, and 3 O. This gives you the empirical formula of ZnCO3, or Zinc Carbonate.

6 0

1 year ago

Plz download the doc and answer its due in 20 mins like bro wth

4vir4ik [10]

Answer:

and how do we know this is not a virus

Explanation:

sorry dude im just saying

4 0

2 years ago

Read 2 more answers

A topographical map of a section of Charleston, SC is shown. What feature is located at the dot marked with an X? A) the center

Effectus [21]

It seems that you have missed the given image to answer this question. But anyway, I found it and got the answer. Based on the topographical map of a section of Charleston, SC, the feature that is located at the dot marked with an X is the high point of a hill. The answer would be option D.

6 0

3 years ago

A) Compute the repeat unit molecular weight of polystyrene. B) Compute the number-average molecular weight for a polystyrene for

Andrej [43]

Answer:

Explanation:

In Polystrene, the molecular formula for the repeat unit = $C_8H_8$ ;

and the atomic weights of Carbon C = 12.01 g/mol

For Hydrogen, it is 1.01 g/mol

Hence, the repeat unit molecular weight is:

m = 8 (12.01 g/mol)+8(1.01 g/mol)

m = 96.08 g/mol + 8.08 g/mol

m = 104.16 g/mol

The degree of polymerization = no-average molecular weight/repeat unit molecular weight.

Mathematically;

$DP = \dfrac{\overline M_n}{m}$

$\overline M_n= DP \times m$

$\overline M_n= 25000 \times 104.16 \ g/mol$

$\overline M_n= 2604000 \ g/mol$

7 0

2 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago