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kozerog [31]
1 year ago
15

What has a higher lattice energy? MgS or BaF2

Chemistry
1 answer:
irinina [24]1 year ago
3 0

Answer:

The lattice energy of MgS is 4 times larger than that of NaCl

Explanation:

because, MgS has a higher charge that is Mg2+,S2− M g 2 + , S 2 − than NaCl (Na+,Cl−) ( N a + , C l − ) .

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Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: spe
Murljashka [212]

Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

  • Heat required to raise the temperature of ice from -20 °C to 0 °C

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 \frac{J}{g*C} and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 \frac{J}{g*C} *20 C and solving: Q=420 J

  • Heat required to convert 0 °C ice to 0 °C water

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g*6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}= 3.333 kJ= 3,333 J (being kJ=1,000 J)

  • Heat required to raise the temperature of water from 0 °C to 60 °C

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 \frac{J}{g*C} and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 \frac{J}{g*C} *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

<u><em> The amount of heat to absorb is 6,261 J</em></u>

<u><em></em></u>

3 0
3 years ago
Derek needs a tool that delivers 25.00 mL of a sodium hydroxide solution. What tool would be best for him to use?
3241004551 [841]
He may use a pipette with a volume greater than 25 ml to transfer the solution.
4 0
3 years ago
Please help! Help me solve problems about naming structures with IUPAC rules
lianna [129]
A. The longest carbon chain is eight, and it has two methyl groups attached to carbon three, and a special group attached to carbon five. Its two names could be:

3-dimethyl-5-(1-methylethyl)octane
3-dimethyl-5-isopropyloctane

Both of these are correct. This is an alkane, because it has all single bonds.

B. This has a triple bond contained between carbons 2 and 3, and has a methyl group off carbon 4. The longest chain is 5. It’s name is:

4-methyl-2-pentyne

This is an alkene, because of the double bond.

C. This has a double bond contained between carbons 2 and 3, and has a methyl off of four and an methyl off of six. The longest chain is eight (follow the longest chain of carbons).

4,6-dimethyl-2-octene

This is an alkene, because of the double bond.

D. This has an ethyl group at 1 and a methyl group at 2 (rotate the compound to make it as clean as possible, in this case, the ring is flipped and rotated to make it alphabetical with the smallest numbers possible). The two names are:

1-ethyl-2-methylbenzene
ortho-ethylmethylbenzene

Both are correct, the ortho prefix telling the location of the ethyl and methyl groups. This is an aromatic structure because of its double bonded ring.

E. The longest chain is nine, and has methyls at three, five, and seven, along with a propyl at five. The name is:

3,5,7-trimethyl-5-propylnonane

This is an alkane, due to the single bonds.

Hope this helps!
7 0
3 years ago
Which is the best example of a new substance forming?
Hunter-Best [27]

Answer:

lighting a match is the answer.

8 0
3 years ago
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This element is found in group 1, period 7 and has the lowest mass in this period. *
liraira [26]

The answer is Francium.

3 0
3 years ago
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