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1 year ago

Which pair of molecules has hydrogen bonding interactions?.

Chemistry

1 answer:

Liula [17]1 year ago

3 0

Answer:

Hydrogen bonding is a special type of dipole-dipole interaction that occurs between the lone pair of a highly electronegative atom (typically N, O, or F) and the hydrogen atom in a N–H, O–H, or F–H bond.

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How many of the planets have an orbital period of less than one Earth year?

GrogVix [38]

Since orbital period depends on how far you are from the sun, planets closer to the sun have a orbital period less than one earth year.

These planets are Mercury and Venus

8 0

2 years ago

For a particular reaction, Δ=−111.4 kJ and Δ=−25.0 J/K.

vichka [17]

Answer:

$\Delta G =-103.95kJ$

Explanation:

Hello there!

In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:

$\Delta G =\Delta H-T\Delta S$

It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:

$\Delta G =-111.4kJ-(298K)(-25.0\frac{J}{K}*\frac{1kJ}{1000J} )\\\\\Delta G =-103.95kJ$

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6 0

2 years ago

A radiometer is shown below. When infrared light passes through the radiometer

natita [175]

We can’t see the pic

4 0

2 years ago

Help please !!!!!!!!

Elena L [17]

Answer:

Option B. 2096.1 K

Explanation:

Data obtained from the question include the following:

Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹

Entropy (S) = +614 JK¯¹mol¯¹

Temperature (T) =.?

Entropy is related to enthalphy and temperature by the following equation:

Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)

ΔS = ΔH / T

With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:

ΔS = ΔH / T

614 = 1287000/ T

Cross multiply

614 x T = 1287000

Divide both side by 614

T = 1287000/614

T = 2096.1 K

Therefore, the temperature at which the reaction will be feasible is 2096.1 K

5 0

3 years ago

If 1,079.75 Joules of heat are added to 77.75 grams of water, by how many degrees Celsius would the water increase? Assume water

djyliett [7]

Answer:

If 1,079.75 Joules of heat are added to 77.75 grams of water, by 3.32 degrees Celsius the temperature of water will increase

$\Delta T=3.32 C^{0}$

Explanation:

$q = mC\Delta T$

Here , q = heat added / removed from the substance

m = mass of the substance taken

$\Delta T$ = Change in temperature

C = specific heat capacity of the substance

In liquid state the value of C for water is :

$4.18 J/gC^{0}$

Given values :

q = 1079.75 J

m = 77.75 gram

Insert the value of C, m , q in the given equation

$q = mC\Delta T$

on transposing ,

$\Delta T=\frac{q}{mC}$

$\Delta T=\frac{1079.75}{77.75\times 4.18}$

$\Delta T=3.32 C^{0}$

3 0

3 years ago