Gold ions occur in seawater to the extent of1.0.

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

$\Rightarrow T_{2}=260K$

$\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}$

So at -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

5 0

3 years ago

Calculate the mass of 6 moles of H2O

RUDIKE [14]

6mole of H2O contains 18×6 g=108 g

8 0

3 years ago

Read 2 more answers

You are performing a chemical reaction in a test tube. The test tube gets colder as the reaction takes place. This chemical reac

Liula [17]

Answer: Option (b) is the correct answer.

Explanation:

A chemical reaction in which heat energy is absorbed by the reactant molecules is known as an endothermic reaction.

Therefore, upon completion of this reaction the container in which reaction is carried out becomes colder.

A chemical reaction in which heat energy is released by the reactant molecules is known as an exothermic reaction.

Therefore, upon completion or during this type of reaction the container in which reaction is carried out becomes hot.

Thus, we can conclude that when the test tube gets colder as the reaction takes place then it means this chemical reaction is endothermic reaction.

4 0

3 years ago

A reaction mixture at 175 k initially contains 522 torr of no and 421 torr of o2. at equilibrium, the total pressure in the reac

zzz [600]

The chemical equation would be:

2NO(g) + O2(g) --> 2NO2 (g)

At equilibrium state, the partial pressure of the gases would be as follows : 

NO = 522 - 2x 

O2 = 421 - x 

NO2 = 2x 
- - - - - - - - - - - - -
943 - x = 748 

x = 195

Calculating for Kp,

Kp = (NO2)^2/ ((NO)^2 * (O2)) 

Kp = (2 * 195)^2/ ((522 - 2 * 195)^2 * (421 - 195)) 

Kp = 0.0386

4 0

3 years ago

An organic acid was analyzed and the following percent composition was obtained: 68.85% carbon, 4.9% hydrogen, and 26.2% oxygen.

rewona [7]

Answer:

The molecular formula of the compound is: C14H12O4

Explanation:

Step 1: find molality

ΔT = Kf x m

ΔT = Tfinal - Tinitial = 3.37 - 5.50 = -2.13

molality (m) = ΔT/Kf = -2.13/-5.12 = 0.416 m

step 2: find number of moles

m = number of moles (n)/Kg solvent which is the benzene

n = molality x Kg solvent = 0.416 molality X (10 x 10^-3 Kg) = 0.00416 mole

Step 3: find molecular mass of the organic compound

n = mass/Molecular mass = m/Mm

Mm = m/n = 1.02/0.00416 = 245.19 g/mole of organic compound molecular formula

Step 4: find number of mole of each compound

let consider it being in 100g where each % will correspond to the mass of the element in 100g of the compound.

C = mass/molar mass = 68.85/12.01 = 5.7327 mole

H = mass/molar mass = 4.9/1.01 = 4.8515 mole

O = mass/molar mass = 26.2/16 = 1.6375 mole

Step 5: divide by smallest number of mole

C = 5.7327/1.6375 = 3.5 = 7/2

H = 4.8515/1.6375 = 3

O = 1.6375/1.6375 = 1

C7/2H3O1

multiply by two to remove the fraction: C7H6O2 in which the empirical molecular mass = 122.13 g/mole

to find the factor for the molecular formula x = Mm of molecular/Mm of empirical = 245.19/122.13 = 2

Multiply the empirical by 2 = 2 x (C7H6O2) = C14H12O4

8 0

3 years ago