All categories

3 years ago

Select the correct answer.

Chemistry

2 answers:

Oduvanchick [21]3 years ago

4 0

Answer:

Ocean temperatures experience high fluctuations. During the day, the land cools the hotter air temperatures above the ocean.

Explanation:

Gwar [14]3 years ago

3 0

Answer:

C) is correct answer . 

You might be interested in

A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas that react to form water vapor. Assume the volume of the con

maxonik [38]

Mole fraction of Oxygen=0.381

Mole fraction of Oxygen= (range of moles of oxygen) ÷(general moles)

also, mole fraction of oxygen = (partial stress of oxygen) ÷ (total strain)

consequently , mole fraction of Oxygen= (2.31 atm)÷(2.31 atm + 3.75 atm)

= 0.381

The mole fraction may be calculated by means of dividing the variety of moles of 1 element of a solution by the entire quantity of moles of all the additives of a solution. It is cited that the sum of the mole fraction of all of the components inside the solution should be identical to 1.

Mole fraction is a unit of awareness. in the solution, the relative amount of solute and solvents are measured by way of the mole fraction and it's far represented through “X.” The mole fraction is the variety of moles of a selected aspect inside the answer divided by way of the entire range of moles in the given answer.

Mole fraction is the ratio between the moles of a constituent and the sum of moles of all ingredients in a mixture. Mass fraction is the ratio between the mass of a constituent and the full mass of a mixture.

The question is incomplete. Please read below to find the missing content.

Assuming that only the listed gases are present, what would the mole fraction of oxygen gas be for each of the following situations? A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas react to form water vapor. Assume the volume of the container and the temperature inside the container does not change.

Learn more about the mole fraction here brainly.com/question/14783710

#SPJ1

4 0

2 years ago

How are distillation and reverse osmosis alike?

Andrei [34K]

I'm having trouble with this one. At the very least, they both purify water. If I'm not mistaken, distillation is part of the process of purifying water.
Really sorry, I hope I helped at least a little! If you have any other questions I might be able to answer better, let me know..

5 0

3 years ago

What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe

Deffense [45]

Answer:

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ C CH3COOH + C CH3COONa = [ CH3COOH ] + [ CH3COO- ]

∴ C CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ C CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒ [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = C CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

7 0

3 years ago

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h

statuscvo [17]

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

6 0

3 years ago

What makes landing the hot air balloon a challenge?

Mariana [72]

You have to think about the fact that hot air is constantly being blown up elevating the balloon once that air is taken away it's up to the steering and the wind after that. I know that probably doesn't help did you want to know the science part?

7 0

3 years ago