What produced the sulfuric acid that helped form the Lechugilla caves

elena55 [62]

I think co 2 maybe
I cant remember

8 0

3 years ago

John dissolves .5g of a white powder in 25g of benzene (FP 5oC) (kf benzene is 5.1) and finds the solution freezes at 3.7oC. Det

navik [9.2K]

Answer:

The compound has a molar mass of 78.4 g/mol

Explanation:

Step 1: data given

Mass of a sample = 0.5 grams

Mass of benzene = 25 grams

Freezing poing = 5 °C

Kf of benzene = 5.1 °C/m

Freezing point solution = 3.7 °C

Step 2: Calculate molality

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 5.0 - 3.7 = 1.3 °C

⇒with i = the can't hoff factor = 1

⇒with Kf = the freezing point depression constant of benzene = 5.1 °C/m

⇒with m = the molality

1.3 = 5.1 * m

m = 1.3 / 5.1

m = 0.255 moles /kg

Step 3: Calculate moles

Molality = moles / mass benzene

0.255 molal = moles / 0.025 kg

Moles = 0.255 molal * 0.025 kg

Moles = 0.006375 moles

Step 4: Calculate molar mass of the compound

Molar mass compund = mass / moles

Molar mass compound = 0.5 grams / 0.006375 moles

Molar mass compound = 78.4 g/mol

The compound has a molar mass of 78.4 g/mol

7 0

2 years ago

Send Message

Ganezh [65]

Answer:

1 Atm

Explanation:

Dalton's law

The total pressure is 3 Atm so all you have to do is subtract the other partial pressures from 3

4 0

3 years ago

Read 2 more answers

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$

$molH_{2} O = 0,783 mol$

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴ $MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}$

$MassOfH_{2}O = 14,09g$

5 0

3 years ago

A man sits on the grass eating a doughnut and sipping iced tea. The sun is a glowing

arsen [322]

Solid liquid gas or plasma. Solid would be ice cubes (frozen state), liquid would be the water inside of the glass, the gas would be when you take that same water and place it on a stovetop to bring to a boiling point. (Gas state) and finally the plasma state is when you your gas particles (from the gas state) if you heat the gas particles, they will turn to plasma.

4 0

1 year ago