What is the hardest metal?

How many calories of heat are necessary to raise the temperature of 319.5 g of water from 35.7 °C

rjkz [21]

20600Cal

Explanation:

Given parameters:

Mass of water = 319.5g

Initial temperature = 35.7°C

Final temperature = 100°C

Unknown:

Calories needed to heat the water = ?

Solution:

The calories is the amount of heat added to the water. This can be determined using;

H = m c Ф

c = specific heat capacity of water = 4.186J/g°C

H is the amount of heat

Ф is the change in temperature

H = m c (Ф₂ - Ф₁)

H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J

Now;

1kilocalorie = 4184J

85996.56J to kCal; $\frac{85996.56}{4184}$ = 20.6kCal = 20600Cal

learn more:

Specific heat brainly.com/question/3032746

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6 0

3 years ago

Partner Namets) For each of the following reactions carried out: Write the balanced chemical equation, the full ionic equation,

torisob [31]

Answer : The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

Reactants are lead nitrate and potassium iodide.

Products are lead iodide and potassium nitrate.

The spectator ions are, $K^+,NO_3^-$

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When potassium iodide react with lead nitrate then it gives potassium nitrate and lead iodide as a product.

The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

The ionic equation in separated aqueous solution will be,

$2K^+(aq)+2I^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^{-}(aq)$

In this equation, $K^+\text{ and }NO_3^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Pb^{2+}(aq)+2I^{-}(aq)\rightarrow PbI_2(s)$

4 0

3 years ago

Iron(III) sulfide reacts with hydrochloric acid to produce hydrogen sulfide and iron(II) chloride. Wright the formula equation f

NeX [460]

Answer:

Fe2S3 + 4HCl —> 2H2S + 2FeCl2 + S

Explanation:

From the question, we were told that:

Iron(III) sulfide reacts with hydrochloric acid to produce hydrogen sulfide and iron(II) chloride.

Please note that elemental sulphur is produced along with this reaction. The equation for the reaction is given below:

Fe2S3 + HCl —> H2S + FeCl2 + S

Now, we can balance the equation as shown below:

There are 3 atoms of S on the left side of the equation and 2 atoms on the right. It can be balance by putting 2 in front of H2S as shown below:

Fe2S3 + HCl —> 2H2S + FeCl2 + S

Now, we have 4 atoms of H on the right side and 1atom on the left side. It can be balance by putting 4 in front of HCl as shown below:

Fe2S3 + 4HCl —> 2H2S + FeCl2 + S

Also, there are 4 atoms of Cl on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of FeCl2 as shown below:

Fe2S3 + 4HCl —> 2H2S + 2FeCl2 + S

Now, we can see that the equation is balanced

5 0

3 years ago

What is the ratio by atoms of elements present in hafnium phosphite?

timurjin [86]

3:6:2 is the correct answer

3 0

3 years ago

What is the hybridization of the central iodine atom in i3−?

Ainat [17]

The central iodine atom in triiodide has sp3d hybridization.In triiodide anion, the central iodine atom has three equatorial lone pairs of electrons and the terminal iodines are bonded axially in a linear shape. Electrons in sp3d hybridization are arranged in trigonal bipyramidal symmetry.
Explanation:
When you draw the Lewis structure of this particle, you'll realize that the central I atom has a pair of bonds and three individual pairs of electrons. as a result of there are five things around that central I atom, it's sp3d hybridized.

The bonds during a gas (CH4) molecule are fashioned by four separate however equivalent orbitals; one 2s and 3 2p orbitals of the carbon interbreed into four sp3 orbitals. within the ammonia molecule (NH3), 2s and 2p orbitals produce four sp3hybrid orbitals, one among that is occupied by a lone try of electrons.

8 0

3 years ago

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