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3 years ago

Consider this chemical equation:

Chemistry

2 answers:

11Alexandr11 [23.1K]3 years ago

8 0

Answer:

the answer is very correct as AgNO3 and Cu(NO3)2

scoundrel [369]3 years ago

4 0

Answer:

A,C

Explanation:

the person above me is correct

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Benzene has a specific gravity of 0.88. if it was spilled in a river what would it do?

Gelneren [198K]

As we know,
Density of Benzene = 876 Kg/m³
And,
Density of Water = 997 Kg/m³
So,
Specific Gravity is calculated as,

Specific Gravity = Density of Benzene / Density of Water

Specific Gravity = 876 Kg/m³ / 997 Kg/m³

Specific Gravity = 0.878

Every object having specific gravity less than 1 will float on water and if value is greater than 1 then it will sink.

Benzene being non-polar in nature does not mix with water and due to less density it will float on the surface of water.

4 0

3 years ago

Explain why substances with a lower boiling point are removed near the top of a petroleum fractional distillation tower.

Anon25 [30]

The lighter components are able to rise higher in the column before they are cooled to their condensing temperature, allowing them to be removed at slightly higher levels.

I hope this helps

6 0

3 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

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 $CO_2=13.00g$

Mass of $H_2O=2.662g$

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 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

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 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $CHO_2$

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 = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

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

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

3 years ago

What is the value for AG at 5000 K if AH = -220 kJ/mol and S= -0.05 kJ/(mol-K)?

Serhud [2]

Answer:

C. 30 kJ

Explanation:

Hello there!

In this case, in agreement to the thermodynamic definition of the Gibbs free energy, in terms of enthalpy of entropy:

$\Delta G= \Delta H-T\Delta S$

It is possible to calculate the required G by plugging in the given entropy and enthalpy as shown below:

$\Delta G=-220kJ/mol-5000K*-0.05kJ/mol*K\\\\\Delta G=30kJ/mol$

Therefore, the answer is C. 30 kJ .

Best regards!

4 0

2 years ago

An aqueous solution of ____ will produce a basic solution.

Anastasy [175]

NaClO, K2CO3, and NaF will form basic soltions.

6 0

3 years ago

Read 2 more answers