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2 years ago

How would the tides on Earth be affected by the absence of the Moon?There would be no spring tides.

Chemistry

2 answers:

Alex787 [66]2 years ago

8 0

Answer:

There are two correct answers:

There would be no spring tides.

The tides would occur at a much lower intensity.

Explanation:

Dmitry_Shevchenko [17]2 years ago

4 0

I think the best answer is: The tides would occur at a much lower intensity.
The forces that create tides include centrifugal forces from the earths rotation and also the gravitational force of the sun. Although these are much smaller than the force exerted by the moon we would still have tides.

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How many moles are there in 3.612 X 10^24 molecules of phosgene (COCL2)

Ugo [173]

Answer:

Explanation:

No of molecules = 3.612*10(24)

Avogadro's number = 6.022*10(23)

no of moles = No of molecules/Avogadro's number

No of moles =3.612*10(24)/6.022*10(23)

No of moles = 6mol

3 0

2 years ago

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Identify the compounds below as a brønsted-lowry acid or lewis acid by clicking and dragging it into the correct column

Vesna [10]

There are many types of acid or bases. Based on the Bronsted-Lowry definition,

* A Bronsted-Lowry acid is a proton donor
* A Bronsted-Lowry base is a proton acceptor

Take this reaction for example:
HCl(aq)+ NH₃(aq)→NH⁴⁺(aq)+Cl⁻(aq)

HCl donates a proton, so it is a Bronsted-Lowry acid. Consequently, ammonia accepts this proton, so it is the Bronsted-Lowry base.

3 0

2 years ago

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A column is filled with four different liquids of different densities. A red liquid, a blue liquid, a green liquid, and a purple

Svet_ta [14]

Answer:

The order of liquid from top to bottom is as follow

Purple, green, red, blue

Explanation:

Chart of densities:

Red = 1.2 g/cm∧3
Blue = 1.6 g/cm∧3
green = 0.8 g/cm∧3
Purple = 0.1 g/cm∧3

Density:

Density is the ratio of mass and volume as follow

d = m/v

Summary:

Greater the density, it will be at bottom and vice versa.
Blue liquid has greater density so it will be at bottom
Purple liquid has low density, so it will be at top.

5 0

2 years ago

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If a 0.4856 gram sample of khp is dissolved in sufficient water to prepare 250 ml of solution, and 25 ml of the solution require

dangina [55]

Mass of potassium hydrogen pthalate KHP is 0.4856 g, its molar mass is 204.22 g/mol, number of moles of KHP can be calculated as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass, putting the values,

$n=\frac{0.4856 g}{204.22 g/mol}=0.00237 mol$

This will be number of moles of NaOH at equivalent point.

Detailed calculations:

Molarity is defined as number of moles in 1 L of solution, for 250 mL of solution, molarity will be:

$M=\frac{0.00237 mol}{250 \times 10^{-3}L}=0.009511 M$

For 25 mL, apply dilution law as follows:

$M_{1}V_{1}=M_{2}V_{2}$

Putting the values,

$0.009511\times 250=M_{2}\times 25 mL$

On rearranging,

$M_{2}=\frac{0.009511\times 250}{25}=0.09511 M$

Convert molarity into number of moles,

$n=M\times V=0.09511 mol/L\times 25\times 10^{-3}L=0.00237 mol$

At equivalent point, number of moles of KHP will be equal to NaOH, thus, number of moles of NaOH will be 0.00237 mol.

Calculation for molarity:

Volume of NaOH is 18.75 mL, thus, molarity can be calculated as follows:

$M=\frac{n}{V}$

Putting the values,

$M=\frac{0.00237 mol}{18.75\times 10^{-3}L}=0.1264 M$

Therefore, molarity of NaOH is 0.1264 M

7 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

Read 2 more answers