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Assoli18 [71]
3 years ago
10

How many grams of calcium hydroxide are required to completely neutralize 467 mL of 0.43 M

Chemistry
1 answer:
podryga [215]3 years ago
8 0

Answer:

456g

Explanation:

that's the answer for the question

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El aguacate al oxidarse, ¿que fenómeno es?, ¿físico, químico o alotropicos?​
devlian [24]

Explanation:

WE A është ai Në The H Përshëndetje Nme është

3 0
3 years ago
Need this ASAP!
sukhopar [10]

Answer:

B. Lower than 100 °C because hydrogen sulfide has dipole-dipole interactions instead of hydrogen bonding.

Explanation:

Intermolecular bonds exists between seperate molecules or units. Their relative strength determines many physical properties of substances like state of matter, solubility of water, boiling point, volatility, viscosity etc. Examples are Van der waals forces, hydrogen bonds and crystal lattice forces.

In hydrogen sulfide, the intermolecular bond is a dipole-dipole attraction which is a type of van der waals attraction. It occurs as an attraction between polar molecules. These molecules line such that the positive pole of one molecule attracts the negative pole of another.

In water, the intermolecular bond is hydrogen bonds in which an electrostatic attraction exists between the hydrogen atom of one molecule and the electronegative atom of a neighbouring molecule.

Based on their relative strength:

  Van der Waals forces < Hydrogen bonding forces < crystal lattice

This makes water boil at a higher temperature than hydrogen sulfide.

6 0
3 years ago
Draw the location of the epicenter of an earthquake in Carmona, Silang,GMA Cavite if the hypothetical distance of epicenter in C
stellarik [79]

The kinematics and geometry can locate the point where the earthquake occurs: Point of interception of the three circles

The earthquakes occur due to a sudden movement of two tectonic plates, this movement produces longitudinal and transverse waves with different speeds, called S and P waves. The speed of the waves is constant, so we can use the uniform kinematic movement relationship

           v = d / t

Where v is the speed of the wave, d is the distance and t is the time.

The speed of the two types of waves is known, they depend on the material and type of wave varies between 4 and 7 km / S, the time is measured in the seismometers, for which we can calculate the distance.

For a seismometer, the calculated distance corresponds to a circle, so to locate the epicenter of the earthquake we must have a minimum of three earthquakes, the point of intersection of the circles is the place of occurrence of the event, see the attachment for a diagram of the process.

In conclusion, with kinematics and geometry we can locate where the earthquake occurs: Point of interception of the three circles

Learn more about Earthquakes here

brainly.com/question/11446551

4 0
3 years ago
Why did scientists need to establish an international language for elements
Ulleksa [173]
So it could be used in every country(different languages) yet still understood
5 0
3 years ago
At 298 K the standard enthalpy of combustion of sucrose is -5645 kJ/mol and the standard reaction Gibbs energy is -5798 kJ/mol.
natka813 [3]

Explanation:

The given data is as follows.

             T = 298 K,          \Delta H^{o} = -5645 kJ/mol

          \Delta G^{o} = -5798 kJ/mol

Relation between \Delta H and \Delta G are as follows.

          \Delta G^{o} = \Delta H^{o} - T \Delta S^{o}    

             -5798 kJ/mol = -5645 kJ/mol - 298 \times \Delta S^{o}

                       -153 kJ/mol = -298 \times \Delta S^{o}

                    \Delta S^{o} = 0.513 kJ/mol K

Now, temperature is 37^{o}C = (37 + 273) K = 310 K

Since,        \Delta G = \Delta H^{o} - T \Delta S^{o}

                            = -5645 kJ/mol - 310 K \times 0.513 kJ/mol K

                            = (-5645 kJ/mol - 159.03 kJ/mol)

                            = -5804.03 kJ/mol

As, change in Gibb's free energy = maximum non-expansion work

            \Delta G = \Delta G_{310 K} - \Delta G_{298 K}

                           = -5804.03 kJ/mol - (-5798 kJ/mol)

                           = -6.03 kJ/mol

Therefore, we can conclude that the additional non-expansion work is -6.03 kJ/mol.

5 0
3 years ago
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