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2 years ago

How are exothermic and endothermic reactions linked in the process of refining metal ore?

1 answer:

6 0

The intended sense is that of a reaction that depends on absorbing heat if it is to proceed. The opposite of an endothermic process is an exothermic process, one that releases "gives out" energy in the form of heat

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Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution accord

Marina CMI [18]

Answer:

$M_{acid}=0.563M$

Explanation:

Hello there!

In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, we solve for the molarity of the acid to obtain:

$M_{acid}=\frac{M_{base}V_{base}}{V_{acid}} \\\\ M_{acid}=\frac{33.98mL*0.1656M}{10.0mL}\\\\ M_{acid}=0.563M$

Regards!

5 0

2 years ago

How many moles of O2 should be supplied to burn 1 mol of C3H8 (propane) molecules in a camping stove

Shkiper50 [21]

The combustion of any hydrocarbon yields water and carbon dioxide. We will now construct a balanced equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Each mole of propane requires 5 moles of oxygen.

4 0

2 years ago

Arranges the following molecules in order of increasing dipole moment: <br> H2O, H2S, H2Te, H2Se.

erastova [34]

Explanation:

Dipole moment is defined as the measurement of the separation of two opposite electrical charges.

$H_{2}O$ is a bent shaped molecule with a dipole moment of 1.87.

$H_{2}S$ is also a bent shaped molecule with a dipole moment of 1.10.

$H_{2}Te$ is a also a bent shaped molecule and has a negligible dipole moment.

$H_{2}Se$ has a dipole moment of 0.29.

Therefore, given molecules are arranged according to their increasing dipole moment as follows.

$H_{2}Te$ < $H_{2}Se$ < $H_{2}S$ < $H_{2}O$

7 0

2 years ago

In the PhET simulation, make sure that the checkbox Stable/Unstable in the bottom right is checked. Using the PhET simulation as

lord [1]

Answer:

kindly check the EXPLANATION SECTION

Explanation:

In order to be able to answer this question one has to consider the neutron proton ratio. Considering this ratio will allow us to determine the stability of a nuclei. The most important rule that helps us in determination of stability is that when the Neutron- Proton ratio of any nuclei ranges from to 1 to 1.5, then we say the nuclei is STABLE.

Also, we need to understand that when the Neutron- Proton ratio is LESS THAN 1 or GREATER THYAN 1.5, then we say the nuclei is UNSTABLE.

So, let us check which is stable and which is unstable:

a. 4 protons and 5 neutrons = Neutron- proton ratio = N/P = 5/4= stable.

b. 7 protons and 7 neutrons = Neutron- proton ratio = N/P = 7/7= 1 = stable.

c. 2 protons and 3 neutrons = Neutron- proton ratio = N/P = 3/5 =0.6 =unstable.

d. 3 protons and 0 neutrons = Neutron- proton ratio = N/P = 0/3= 0= unstable.

e. 6 protons and 5 neutrons = Neutron- proton ratio = N/P = 5/6= 0.83 = unstable.

f. 9 protons and 9 neutrons = Neutron- proton ratio = N/P = 9/9 = 1 = stable.

g. 8 protons and 7 neutrons = Neutron- proton ratio = N/P = 7/8 =0.875 = unstable.

h. 1 proton and 0 neutrons = Neutron- proton ratio = N/P = 0/1 =0 = unstable

6 0

2 years ago

Metals with ________ electron configurations characteristically form diamagnetic, square planar complexes.

erastovalidia [21]

Answer:

Explanation:

Diamagnetism , paramagnetism and ferromagnetism are terms which are used to describe the magnetic activity of transition metals. These terms helps to know the response the metal will have when placed in a magnetic field. These activities can be discerned from the d-block electronic configuration. If the number of electrons are even, this means they all form a pair and the metal is diamagnetic. If otherwise, there is an unpaired electron which causes the paramagnetic activity of the metal in magnetic field.

To the question, options A-D are diamagnetic but configurations with d8 are the ones that form square planar complexes

8 0

2 years ago