All categories

2 years ago

Below is a word equation. Please follow directions for each part below.

Chemistry

1 answer:

Airida [17]2 years ago

7 0

Answer:

C2H6 + 02 ---> CO2 + H20

balanced equation -

C2H6 + 7/202 ---> 2CO2 + 3H20

Explanation:

for equation to be balanced, we must have the same amount of each element on both sides. hence the numbers to balance the equation are 7/2, 2 and 3 respectively.

You might be interested in

What type of galaxy do we live in

strojnjashka [21]

We live in the Milky Way Galaxy

7 0

2 years ago

Why do you think the molarity relationship is so important to the chemist?

Oksana_A [137]

There are two big advantages of using molarity to express concentration. The first advantage is that it's easy and convenient to use because the solute may be measured in grams, converted into moles, and mixed with a volume.

The second advantage is that the sum of the molar concentrations is the total molar concentration. This permits calculations of density and ionic strength

5 0

2 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

2 years ago

What is the molarity of 225 grams of Cu(NO2)2 in a total volume of 2.59 L?

Kay [80]

Answer:

The molarity is 0.56 $\frac{moles}{L}$

Explanation:

In a mixture, the chemical present in the greatest amount is called a solvent, while the other components are called solutes. Then, the molarity or molar concentration is the number of moles of solute per liter of solution.

In other words, molarity is the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by:

$Molarity (M)=\frac{number of moles of solute}{Volume}$

Molarity is expressed in units ( $\frac{moles}{liter}$ ).

Then you must know the number of moles of Cu(NO₂)₂. For that it is necessary to know the molar mass. Being:

Cu: 63.54 g/mol
N: 14 g/mol
O: 16 g/mol

the molar mass of Cu(NO₂)₂ is:

Cu(NO₂)₂= 63.54 g/mol + 2*(14 g/mol + 2* 16 g/mol)= 155.54 g/mol

Now the following rule of three applies: if 155.54 g are in 1 mole of the compound, 225 g in how many moles are they?

$moles=\frac{225 g*1 mole}{155.54 g}$

moles= 1.45

So you know:

number of moles of solute= 1.45 moles
volume=2.59 L

Replacing in the definition of molarity:

$Molarity=\frac{1.45 moles}{2.59 L}$

Molarity= 0.56 $\frac{moles}{L}$

The molarity is 0.56 $\frac{moles}{L}$

5 0

3 years ago

The vapor pressure of water at 25.0°C is 23.8 torr. Determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.

mrs_skeptik [129]

According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.

the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent

$\frac{p^{0-}p}{p^{0}}=x_{B}$