Please help!

4.Which sentence BEST describes the relationship between the plates in Earth's crust and the sizes of the landmasses on top of t

riadik2000 [5.3K]

Answer:

D. The plates are different in size from the landmasses.

Explanation:

The tectonic plates that are also called lithospheric plates are very massive and irregular-shaped rocks that are composed of both the continental and the oceanic lithosphere and they may vary from a few hundred to thousands of square km.

5 0

3 years ago

What is litmus paper used for? To indicate if all the reactants have been used up to determine if a chemical reaction has occurr

rjkz [21]

The main use of litmus is to test whether a solution is acidic or basic. Blue litmus paper turns red under acidic conditions and red litmus paper turns blue under basic or alkaline conditions, with the color change occurring over the pH range 4.5–8.3 at 25 °C (77 °F).

8 0

3 years ago

A 100.0g sample of tin is heated to 100.0 oC (Celsius) and is placed in a coffee cup calorimeter containing 150. g of water at 2

Snowcat [4.5K]

Explanation:

It is known that specific heat of water is 4.184 $J/g^{o}C$ and atomic mass of tin is 118.7 g/mol. For the given situation,

$Q_{lost} = Q_{gained}$

Let us assume that,

$m_{1}$ = mass of Sn

$m_{2}$ = mass of $H_{2}O$

Therefore, heat energy expression for heat lost and gained is as follows.

$Q_{lost} = Q_{gained}$

$m_{1}C_{1}(T_{2} - T_{1}) = m_{2}C_{2}(T_{1} - T_{2})$

$100 g \times C_{1} \times (100^{o}C - 27.4^{o}C) = 150 g \times 4.184 /g^{o}C \times (27.4^{o}C - 25^{o}C)$

$7260C_{1} = 150 \times 4.184 \times 2.4$

$C_{1} = \frac{1506.24}{7260}$

= 0.207 $J/g^{o}C$

For, 118.7 g the specific heat of tin will be calculated as follows.

$C_{1} = 0.207 J/g^{o}C \times 118.7 g$

= 24.5 $J/mol^{o}C$

Thus, we can conclude that specific heat of tin is 24.5 $J/mol^{o}C$ .

3 0

3 years ago

If you gently shake a carbon dioxide fire extinguisher, you will feel the presence of liquid within the extinguisher What condit

scoundrel [369]

When we wish to convert a gas to liquid we have to either

a) decrease temperature

b) increase pressure

In case of fire extinguisher the CO2 is found to be in liquid state, this is as the CO2 is pressurized at high pressure which keeps CO2 in liquid state

the ideal pressure and temperature conditions when CO2 gas can be converted to CO2 gas

Pressure = 5 - 73 atm

Temperature = -57 to 31 degree Celsius

6 0

3 years ago

The key step in the metabolism of glucose for energy is the isomerization of glucose−6−phosphate (G6P) to fructose−6−phosphate (

vlabodo [156]

Answer :

(A) The value of $\Delta G_{rxn}$ is, 7.37 kJ/mol

(B) The value of $\Delta G_{rxn}$ is, -4.03 kJ/mol

(C) The value of Q for the reaction is, 1.39

Explanation :

The given balanced chemical reaction is,

$G6P\rightarrow F6P$

The expression for reaction quotient will be :

$Q=\frac{[F6P]}{[G6P]}$

Given:

$Q=\frac{[F6P]}{[G6P]}=10.0$

k = 0.510

First we have to calculate the $\Delta G^o$ for the reaction.

$\Delta G^o=-RT\ln k$

$\Delta G^o=-(8.314J/mol.K)\times (298K)\times \ln (0.510)$

$\Delta G^o=1668.259J/mol=1.67kJ/mol$

Part A:

Now we have to calculate the value of $\Delta G_{rxn}$ .

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = ?

$\Delta G_^o$ = standard Gibbs free energy = 1.67 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

Q = reaction quotient = 10.0

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(1.67kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln (10.0)$

$\Delta G_{rxn}=7.37kJ/mol$

Thus, the value of $\Delta G_{rxn}$ is, 7.37 kJ/mol

Part B:

Now we have to calculate the value of $\Delta G_{rxn}$ .

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = ?

$\Delta G_^o$ = standard Gibbs free energy = 1.67 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

Q = reaction quotient = 0.100

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(1.67kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln (0.100)$

$\Delta G_{rxn}=-4.03kJ/mol$

Thus, the value of $\Delta G_{rxn}$ is, -4.03 kJ/mol

Part C:

Now we have to calculate the value of Q.

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = 2.50 kJ/mol

$\Delta G_^o$ = standard Gibbs free energy = 1.67 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

Q = reaction quotient = ?

Now put all the given values in the above formula 1, we get:

$2.50kJ/mol=(1.67kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln (Q)$

$Q=1.39$

Thus, the value of Q for the reaction is, 1.39

8 0

4 years ago