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2 years ago

How does gravity depend on the mass of the two objects

Chemistry

1 answer:

marin [14]2 years ago

3 0

Answer:

Yes

Explanation:

the gravity of an object depends on its mass. that's why the sun has its gravitational pull.

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Lithium has two stable isotopes with masses of 6.01512 amu and 7.01600 amu. The average molar mass of Li is 6.941 amu. What is t

Dvinal [7]

Answer : The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % respectively.

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'

For Li isotope-1 :

Mass of Li isotope-1 = 6.01512 amu

Fractional abundance of Li isotope-1 = x

For Li isotope-2 :

Mass of Li isotope-2 = 7.01600 amu

Fractional abundance of Li isotope-2 = 100-x

Average atomic mass of Li = 6.941 amu

Putting values in equation 1, we get:

$6.941=[(6.01512\times x)+(7.01600\times (100-x))]$

By solving the term 'x', we get:

$x=694.048$

Percent abundance of Li isotope-1 = $\frac{694.048}{100}=6.94\%$

Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %

6 0

3 years ago

What type of cryptography uses two keys instead of just one, generating both a private and a public key?

Alexxx [7]

Answer:

It is known as asymmetric key cryptography it is also called public key cryptography.

Explanation:

Asymmetric key cryptography method makes use of two keys.One is used for encryption and the second one for decryption. The public key serves to encrypt plain text or verify a digital signature, while the private key is used to decrypt or decipher the encrypted text or to create a digital signature.

3 0

3 years ago

A voltaic cell utilizes the following reaction: 4Fe2+(aq)+O2(g)+4H+(aq)→4Fe3+(aq)+2H2O(l). What is the emf of this cell under st

Leviafan [203]

Answer:

0.46 V

Explanation:

The emf for the cell is given by:

Eº cell = Eº oxidation + Eº reduction

From the given balanced chemical equation, we can deduce that Fe²⁺ has been oxidized to Fe³⁺, and O reduced from 0 to negative 2, according to the half cell reactions:

4Fe²⁺ ⇒ Fe³⁺ + 4e⁻ oxidation

O₂ + 4H⁺ + 4 e⁻ ⇒ 2 H₂O reduction

From reference tables for the standard reduction potential, we get

Eº red Fe³⁺ / Fe²⁺ Eºred = 0.77 V

Eº red O₂ / H₂O Eºred = 1.23 V

Now all we need to do is change the sign of Eº reduction for the species being oxidized ( Fe²⁺ ) and add it to Eº reduction O₂:

Eº cell = Eº oxidation + Eº reduction = - (0.77 V ) + 1.23 V = 0.46 V

5 0

4 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

8 0

3 years ago

Paper towel absorbs water is a physical or chemical change?

Marina CMI [18]

It depends. many of the websites i see answering the questions are all over the place, but good luck.

8 0

3 years ago