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Yakvenalex [24]
3 years ago
6

What is true when an element is oxidized? It bonds with the hydroxide ion. It loses electrons to another element. It reacts with

oxygen gas. It takes electrons from another element.
Chemistry
1 answer:
Mama L [17]3 years ago
8 0

Answer: It loses electrons to another element.

Explanation:- Oxidation is the process in which an element loses electrons and there is an increase in the oxidation state. On losing electrons it combines with a electronegative element such as oxygen, sulphur or nitrogen etc.

Fe\rightarrow Fe^{2+}+2e^-

Reduction is the process in which an element gains electrons and there is a decrease in the oxidation state.

\frac{1}{2}O_2+2e^-\rightarrow O^{2-}

You might be interested in
Compute the molar enthalpy of combustion of glucose (C6 H12O6 ): C6 H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O (g) Given that com
lana66690 [7]

Answer:

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

Combustion of 0.305 grams causes a raise of 6.30 °C

Calorimeter has a heat capacity of 755 J/°C

Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g)

Step 3:

ΔH = (m * C * ΔT + c(calorimeter) * ΔT)

with m = mass of the solutin = 0.305 grams

with C = heat capacity of water = 4.184 J/g°C

with ΔT = the change in temperature = 6.30 °C

with c(calorimeter) = 755 J/°C

ΔH = 0.305 * 4.184 *6.30 + 755 * 6.30  = 4764.5 J ( negative because it's exothermic)

Step 4: Calculate moles of glucose

Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = -4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = -2819.3 kJ/moles

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

5 0
3 years ago
Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is no
nasty-shy [4]

This question is incomplete, the complete question is;

Tonksite is a solid at 300.00K.  At 300.00 K its enthalpy of sublimation is 66.00 kJ/mol. The sublimation pressure at 300.00 K is 5.00 × 10⁻⁴ atm

Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is not a function of temperature.

Answer: the sublimation pressure of the solid at the melting point is  0.3727 atm

Explanation:

Given that;

T1 = 300 K

T2 = 400 K

H_sub = 66 kJ/mol = 66000 J/mol

P1 = 5.00 × 10⁻⁴ atm

p2 = ?

now using the expression

log( p2 / 5.00 × 10⁻⁴ ) = (H_sub / R ×  2.303 ) (( T2 - T1) / T1T2)

now we substitute of given values into the expression

log(p2/p1) = (66000 / 8.314 ×  2.303 ) (( 400 - 300) / 300 × 400 )

p2 = 0.3727 atm

therefore the sublimation pressure of the solid at the melting point is  0.3727 atm

6 0
3 years ago
a sample of gold required 2.1200j of heat to melt it from room temperature, 22.0 degrees celsius to its melting point, 1064.4 de
ElenaW [278]

Mass of Gold = 267.165 ×  0.01552494829

⇒ 4.1477228099

The amount of heat(q) required to raise m grams of a substance-specific C from T1 to T2 is given by

q=m C (T2-T1) ........1

Given : q= 2.1200 J

the initial temperature of gold, T1 = 22.0Celcius

the final temperature of gold, T2 = 1064.4Celcius

specific heat of gold = 0.131

putting values in eq 1:

⇒ 2.1200 = m × 0.131 × (1064.4-22)

⇒ 2.1200 = m × 0.131 × 1042.4

⇒ 2.1200 / 136.5544

⇒ 0.01552494829

Since 1g= 0.01552494829 Pounds

Mass of Gold = 267.165 ×  0.01552494829

⇒ 4.1477228099

Learn more about temperature here: brainly.com/question/11464844

#SPJ9

7 0
1 year ago
Please help, any and all is very much appreciated!
iogann1982 [59]
1. True
2. False
3. False
4. True
5. True
6. True
7. All of the above
8. According to their reactivity with other substances
9. Standard temp and pressure
10. Flush your eyes in the eye flush for 15-20 minutes
3 0
2 years ago
Enter your answer in the provided box. When mixed, solutions of barium chloride, BaCl2, and potassium chromate, K2CrO4, form a y
ira [324]

Answer:

35.42g

Explanation:

Step 1:

The balanced equation for the reaction

BaCl2(aq) + K2CrO4(aq) → BaCrO4(s) + 2KCl(aq)

Step 2:

Determination of the limiting reactant.

It is important to determine which of the reactant is limiting the reaction as the limiting reactant is used to determine the maximum yield of the reaction. The limiting reactant can be determined as follow:

From the balanced equation above,

1 mole of BaCl2 reacted with 1 mole of K2CrO4.

Therefore, 0.7 mole of BaCl2 will also react with 0.7 mol of K2CrO4.

From the above illustration, we can see that it requires a higher amount of K2CrO4 to react with 0.7 mol of BaCl2. This simply means that K2CrO4 is the limiting reactant.

Step 3:

Determination of the number of mole of BaCrO4 produced from the reaction.

The limiting reactant is used in this case.

From the balanced equation above,

1 mole of K2CrO4 produced 1 mole of BaCrO4.

Therefore, 0.14 mole of K2CrO4 will also produce 0.14 mole of BaCrO4.

Step 4:

Converting 0.14 mole of BaCrO4 to grams.

This is illustrated below:

Molar Mass of BaCrO4 = 137 + 52 + (16x4) = 137 + 52 + 64 = 253g/mol

Number of mole BaCrO4 = 0.14 mole

Mass of BaCrO4 =?

Mass = number of mole x molar Mass

Mass of BaCrO4 = 0.14 x 253

Mass of BaCrO4 = 35.42g

Therefore, 35.42g of BaCrO4 is produced from the reaction.

4 0
3 years ago
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