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2 years ago

What water activities might be easier to do in Lake Assal's salty water? What activities could be more difficult?

Chemistry

2 answers:

frosja888 [35]2 years ago

6 0

Swimming would be hard, but since you float on the water, there is no need to swim.

cluponka [151]2 years ago

4 0

Swimming in the river would be easy because the salty water has a higher buoyancy hence one can easily swim and are less likely to sink. Water activities that involve drinking the water will cause immediate dehydration because the body will lose water through osmotic pressure.

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The mechanism for the reaction described by 2N2O5(g) ---> 4NO2(g) + O2(g) is suggested to be (1) N2O5(g) (k1)--->(K-1) NO2

aliina [53]

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2 years ago

In two or more complete sentences describe all of the van der Waals forces that exist between molecules of sulfur

zavuch27 [327]

Answer:

Dipole-Dipole attraction

Explanation:

Dipole-dipole attraction is a type of vander waals forces found in the molecules of sulfur dioxide.

Vander waals forces are weak attractions joining non-polar and polar molecules together. They are of two types:

London dispersion forces which are weak attractions found between non-polar molecules.
Dipole-Dipole attraction are the forces of attraction which exists between polar molecules. Such molecules have permanent dipoles. This implies that the positive pole of one molecule attracts the negative pole of another. This is what happens between the oxygen and sulfur molecules.

3 0

2 years ago

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A bottle containing 1,665 g of sulfuric acid (H2SO4, 98.08 g/mol) was spilled in a laboratory. The emergency spill kit contained

Tamiku [17]

Answer:

A. Yes, there is more than enough sodium carbonate.

Explanation:

Hello,

In this case, based on the given reaction which is:

$H_2SO_4(aq) + Na_2CO_3(s) \rightarrow Na_2SO_4(aq) + CO_2(g) + H_2O(l)$

By stoichiometry, one computes the grams of sodium carbonate that will neutralize 1,665 g of sulfuric acid as shown below:

$1,665gH_2SO_4*\frac{1molH_2SO_4}{98.08gH_2SO_4}*\frac{1molNa_2CO_3}{1molH_2SO_4} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}*\frac{1kgNa_2CO_3}{1000gNa_2CO_3}\\m_{Na_2CO_3}=1.80gNa_2CO_3$

Thus, the available mass is 2.0 kg so 0.2 kg are in excess, therefore: A. Yes, there is more than enough sodium carbonate.

Best regards.

5 0

2 years ago

What is the ph of a solution made by combining 157 ml of 0.35 m nac2h3o2 with 139 ml of 0.46 m hc2h3o2? the ka of acetic acid is

ExtremeBDS [4]

The first step is to calculate the molarity of each compound:
final volume of solution = 157 + 139 = 296 mL
molarity of nac2h3o2 = (157 x 0.35) / 296 = 0.1856 molar
molarity of hc2h3o2 = (139 x 0.46) / 296 = 0.216 molar

Then, we calculate the pH as follows:
pKa of acetic acid = -log(1.75 × 10^-5) = 4.7569
pH = pKa + log ([salt] / [acid])
= 4.7569 + log(0.1856 / 0.216)
= 4.691

6 0

2 years ago

495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F

I am Lyosha [343]

Answer:

Total pressure of the flask is 2.8999 atm.

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

= 0.495 L (1 cm³ = 1 mL = 0.001 L)

Volume of nitrogen (N2) gas = 877 cm3

= 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

= 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature = 25 °C

T = (25°C + 273.15) K

= 298.15 K

Pressure = 114.7 kPa

= 114.700 Pa

Pressure (torr) = 114,700 / 101325

= 1.132 atm

Formula:

PV=nRT (ideal gas equation)

P = pressure

V = volume

R (gas constnt)= 0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

For Oxygen:

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

= 0.560 / 24.47

= 0.0229 moles

For Nitrogen:

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

= 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

= 0.0229 moles + 0.0406 moles

n = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K / 0.536 L

P = 1.554 / 0.536

P = 2.8999 atm

Total pressure in the flask is 2.8999 atm, while assuming the temperature constant.

7 0

2 years ago

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