PLEASE HELP ME IDENTIFY THIS ELEMENT

Write a balanced overall reaction from these unbalanced half-reactions.

ANTONII [103]

The first half reaction is:
Sn+ -> Sn2+

You add 1 electron on the right side:
Sn+ -> Sn2+ + 1e-

The second half reacation is:
Ag+ -> Ag

You add 1 electron to the left side:
Ag+ + 1e- -> Ag

Then, you combine both reactions to get the balanced overall reaction since both half reactions have 1 electrons on the right and left side, respectively. You get:
Sn+ + Ag+ -> Sn2+ + Ag

7 0

3 years ago

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metal weighing 6.98 g was heated to 91.29 °C and then put it into 114.84 mL of water (initially at 24.37 °C). The metal and wate

torisob [31]

Answer:

The specific heat of the metal is 10.93 J/g°C.

Explanation:

Given,

For Metal sample,

mass = 6.98 grams

T = 91.29°C

For Water sample,

volume = 114.84 mL

T = 24.37°C.

Final temperature of mixture = 33.54°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that

water sample temperature changed from 24.37°C to 33.54°C and metal sample temperature changed from 91.29°C to 33.54°C.

We have all values, but, here mass of water is not given. It can be found by using the formula

Density = Mass/Volume

Since, density of water = 1 g/mL

we get, Mass = 114.84 grams.

Since specific heat of water is 4.184 J/g°C.

Now substituting all values in (mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

(6.98)(91.29 - 33.54)(Cp) = (114.84)(33.54 - 24.37)(4.184)

solving, we get,

Cp = 10.93 J/g°C.

the specific heat of the metal is 10.93 J/g°C.

7 0

3 years ago

In Europe, gasoline efficiency is measured in km/L. If your car’s gas mileage is 25.0 mi/gal, how many liters of gasoline would

Deffense [45]

Answer:

$V=23.5L$

Explanation:

Hello,

In this case, it is convenient to compute the car's mileage in km/L as follows:

$25.0\frac{mi}{gal}*\frac{1km}{0.6214mi}*\frac{1gal}{3.78L}=10.64\frac{km}{L}$

In such a way, since the distance is measured to be 250 km, the volume requirement is:

$V=\frac{250km}{10.64kg/L}\\ \\V=23.5L$

Regards.

7 0

3 years ago

I need a help from this question!!

Alik [6]

Answer:

B

Explanation:

Please take a look at the picture attached for the drawings and structures.

C2H4 is a alkene (C-C double bond). When steam (water) is added, it turns into an alcohol, where the double bond breaks and a (-OH) functional group is attached to one of the Carbons. In this case, C2H4 ethene is turned into ethanol.

When an alcohol undergoes oxidation, primary alcohols turn into aldehyde (-CHO) or carboxylic acids (-COOH). Secondary alcohol turns into ketone. Ethanol is a primary alcohol. And since it later reacts with propanol, it can only form carboxylic acid when it oxidizes. The product in this reaction is ethanoic acid.

Carboxylic acid reacts with alcohol to form an ester (-COO-). the -COOH group from acid combines with the -OH group from alcohol to form an ester bond -COO- while eliminating H2O. Therefore, when propanol undergoes esterification with ethanoic acid, propyl ethanoate is produced. It is the answer of B.

7 0

4 years ago

Classify each substituent as ortho, para activators; ortho, para deactivators; or meta deactivators.

tia_tia [17]

The activating groups are; -NH2, -NR2 > -OH, -OR> -NHCOR> -CH3. while the deactivating groups are; -NO2, -CF3> -COR, -CO2R, -SO3H > Halogens.

<h3>Electrophillic substitution</h3>

Owing to the unusual stability of the benzene ring, it does not undergo addition like other unsaturated species, rather benzene does undergo electrophillic substitution.

The substituent groups determine whether the molecule will be activated or deactivated towards electrophilic substitution, the following groups decativate benzene towards electrophilic substitution; -NO2, -CF3> -COR, -CO2R, -SO3H > Halogens. The following groups activate benzene ring towrads electrophilic substitution; -NH2, -NR2 > -OH, -OR> -NHCOR> -CH3.

Learn more about activators and decativators: brainly.com/question/9904853

8 0

3 years ago