The 4 in formula 4H2O is a 1. coefficient 2. subscript 3. superscript 4. binomial

If 9.5 atms of pressure were increased to 25 atms of pressure, what would be the final volume

nordsb [41]

Answer:

The final volume of the gas is 36.1 L.

Explanation:

Given:

Initial pressure of the gas is, $P_1=9.5\ atm$

Final pressure of the gas is, $P_2=25\ atm$

Initial volume of the gas is, $V_1=95\ L$

Final volume of the gas is, $V_2=?$

Here, we shall use Boyle's Law which states that for a process under constant temperature, the pressure of the gas changes inversely with the change in volume.

Here, the pressure is increased. So, the volume of the gas is decreased.

Therefore, as per Boyle's Law:

$P_1V_1=P_2V_2\\9.5\times 95=25V_2\\902.5=25V_2\\V_2=\frac{902.5}{25}\\V_2=36.1\ L$

So, the final volume of the gas is 36.1 L.

4 0

3 years ago

if a molecule of carbon dioxide is involved in a chemical reaction, how is it represented in the chemical equation for the react

DIA [1.3K]

Im mot sure but you could ask my mum

4 0

3 years ago

a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh

Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

$Molarity = M = \frac{1,22 mol solute}{lts solution}$

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

$1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute$

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

$1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution$

With the molecular weight of solute (Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol), we can obtain the mass of solute:

$1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute$

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

$molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal$

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

$%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%$

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: Moles solution = moles solute + moles solvent. First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

$702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent$

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

$Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03$

6 0

3 years ago

A 7.27-gram sample of a compound is dissolved in 250 grams of benzene. The freezing point of this solution is 1.02°C below that

almond37 [142]

Answer:

146 g/mol → option b.

Explanation:

This is a problem about the freezing point depression. The formula for this colligative property is:

ΔT = Kf . m . i

We assume i = 1, so our compound is not electrolytic.

ΔT = Freezing T° of pure solvent - Freezing T° of solution = 1.02 °C

m = molality (mol of solute/kg of solvent)

We convert the grams of solvent (benzene) to kg → 250 g . 1 kg/1000 = 0.250 kg.

We replace → 1.02°C = 5.12°C/mol/kg . mol/ 0.250kg . 1

1.02°C / 5.12 mol/kg/°C = mol/ 0.250kg

0.19922 mol/kg = mol/ 0.250kg

mol = 0.19922 . 0.250kg → 0.0498 mol

molar mass = g/mol → 7.27 g / 0.0498mol = 146 g/mol

5 0

3 years ago

a graduated cylinder (approximate as a regular cylinder) has a radius of 1.045 cm and a height of 30.48 cm. what is the volume o

Luba_88 [7]

Answer: 103.5 cm^3

Explanation:

8 0

3 years ago

The 4 in formula 4H2O is a1. coefficient2. subscript3. superscript4. binomial​

The 4 in formula 4H2O is a1. coefficient2. subscript3. superscript4. binomial