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3 years ago

How are physical and chemical changes similar?

Chemistry

2 answers:

kherson [118]3 years ago

7 0

the answer is A lqmndcboerhbfiewrhbc

Ivan3 years ago

4 0

I believe A. Neither can change the number of atoms of each element that are present

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stoichometry in chemistry!! please someone help i'm begging you, if i fail this then i fail chemistry

Anna71 [15]

Answer:

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4 0

2 years ago

How many grams of copper (I) chloride can be produced from the reaction of 73.5 g of copper (I) oxide with hydrochloric acid acc

natulia [17]

The balanced equation for the reaction is as follows

Cu₂O + 2HCl ---> 2CuCl + H₂O

Molar ratio of Cu₂O to CuCl is 1:2

mass of Cu₂O reacted - 73.5 g

Number of moles of Cu₂O reacted - 73.5 g / 143 g/mol = 0.51 mol

According to the molar ratio,

when 1 mol of Cu₂O reacts then 2 mol of CuCl is formed

therefore when 0.51 mol of Cu₂O reacts then - 2 x 0.51 mol of CuCl is formed

number of CuCl moles formed - 1.02 mol

mass of CuCl formed - 1.02 mol x 99 g/mol = 101 g

mass of CuCl formed is 101 g

6 0

3 years ago

Which two systems are represented in this model?

Mama L [17]

Answer:

I think it's c

Explanation:

4 0

3 years ago

Read 2 more answers

Once a slide is prepared and placed onto the microscope, the magnification and focus need to be altered. Describe how to change

Harlamova29_29 [7]

Answer:

我實際上不知道答案，我只是為了點數而這樣做，哈哈，祝你好運哈哈

Explanation:

我實際上不知道答案，我只是為了點我實際上不知道答案，我只是為了點數而這樣做，哈哈，祝你好運哈哈數而這樣做，哈哈，祝你好運哈哈

5 0

2 years ago

Read 2 more answers

The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at

kherson [118]

Answer: The molecular formula of the compound is $C_4H_{10}O_4$

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

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

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol$

Calculating the molecular formula:

The chemical equation for the combustion of compound 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 $CO_2=36.26g$

Mass of $H_2O=14.85g$

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 36.26 g of carbon dioxide, $\frac{12}{44}\times 36.26=9.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 14.85 g of water, $\frac{2}{18}\times 14.85=1.65g$ of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles$

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 0.659 moles.

For Carbon = $\frac{0.824}{0.659}=1.25\approx 1$

For Hydrogen = $\frac{1.65}{0.659}=2.5$

For Oxygen = $\frac{0.659}{0.659}=1$

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is $C_2H_5O_2$

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 = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

$n=\frac{133.9g/mol}{61g/mol}=2$

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

$C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4$

Hence, the molecular formula of the compound is $C_4H_{10}O_4$

4 0

3 years ago