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3 years ago

Do all metal produce electromagnetic radiation when they are heated in a flame

Chemistry

1 answer:

Ilia_Sergeevich [38]3 years ago

3 0

Answer:

Yes , all metals when placed in the heat produces EM radiations

Explanation:

Due to heat , the electrons mobilized and collapsed and electronic transition happened . the electrons shift from low level to high level and so on in the opposite . The electromagnetic radiation are emitted

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What identifies an ion

bezimeni [28]

Answer:

Identifying whether or not an element is an ion is a very simple process. Identify the charge of the element. ... The number of electrons is equal to the atomic number minus the charge of the atom. Refer to an element with either a positive or negative charge as an ion.

7 0

3 years ago

Dissolution of KOH, ΔHsoln:

swat32

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq) ΔHsoln (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

$\Delta H = \Delta H_{soln} + \Delta H_{neut}$

The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:

$\Delta H = \Delta H_{neut} - \Delta H_{soln}$

Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.

Learn more here:

brainly.com/question/2082986?referrer=searchResults

brainly.com/question/1657608?referrer=searchResults

I hope it helps you!

6 0

2 years ago

When a certain amount of MgF2 is added to water, the freezing point lowers by 3.5° C. What was the molality of the

Leni [432]

The molality of the solution is obtained as 0.63 m.

<h3>What is the freezing point?</h3>

The freezing point is the temperature at which the liquid is converted into solid.

We know that;

ΔT = 3.5° C

K = 1.86° C/m

i = 3

m = ?

Thus;

ΔT = K m i

m = ΔT/K i

m = 3.5° C/ 1.86° C/m * 3

m = 0.63 m

Learn more about freezing point:brainly.com/question/3121416

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6 0

2 years ago

The melting point of H₂O(s) is 0 °C. Would you expect the melting point of H₂S(s) to be 85 °C, 0 °C or -85 °C.? Justify your cho

dimulka [17.4K]

Answer:

-85 °C

Explanation:

O and S are in the same group( Group 16). Since S is below O it's atomic mass is higher than O. So molar mass of H2S is higher than H2O. The strength of Vanderwaal Interactions ( London dispersion forces) increases when the molar mass increases. However, only H2O can form H bonds with each other. This is because electronegativity of O is higher than S and therefore H in H2O has a higher partial positive charge than H of H2S.

H bond dominate among these 2 types of forces so the strength of attractions between molecules is higher in H2O than H2S. Therefore more energy should be supplied for H2O to break inter

molecular forces and convert from solid to liquid state than H2S. So mpt of H2O must be higher than that of H2S.

5 0

2 years ago

Weathering of rock and sedimentation ultimately lead to the formation of___

Inga [223]

Weathering of the rock and sedimentation are decomposition processes. Through time, the minerals in the rocks soften due to pressure and heat. So, they crumble down and reduce in terms of size. Once they do, they become sand or part of the soil. So, the answer is A.

6 0

3 years ago

Read 2 more answers