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Yanka [14]
3 years ago
10

How many significant figures are in 0.003450?

Chemistry
1 answer:
Olenka [21]3 years ago
7 0

Answer:

4

Explanation:

Four significant figures

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Alguien me dice cual sería el nombre de esta formulación organica? se lo agradeceria mucho :(
amm1812

Answer:

sorry i d k for me

Explanation:

4 0
3 years ago
Benzoic acid is in the process of being recrystallized. Your lab partner is in a hurry and after the impure benzoic acid is diss
bazaltina [42]

Answer:

Impurities will be trapped in the crystals of the benzoic acid crystallized in this manner.

Explanation:

After benzoic acid is dissolved in hot water, it should have been allowed to cool gradually before it is transferred into an ice bath.

This gradual cooling will aid the separation of impurities so that when the vessel is now submerged in an ice bath, only pure benzoic acid is recrystalized.

If the vessel is immediately submerged into an ice bath, impurities will be trapped in the crystals of the benzoic acid.

6 0
4 years ago
A 1.67-g sample of solid silver reacted in excess chlorine gas to give a2.21-g sample of pure solid Agcl.The heat given off in t
kotegsom [21]

<u>Given:</u>

Mass of Ag = 1.67 g

Mass of Cl = 2.21 g

Heat evolved = 1.96 kJ

<u>To determine:</u>

The enthalpy of formation of AgCl(s)

<u>Explanation:</u>

The reaction is:

2Ag(s) + Cl2(g) → 2AgCl(s)

Calculate the moles of Ag and Cl from the given masses

Atomic mass of Ag = 108 g/mol

# moles of Ag = 1.67/108 = 0.0155 moles

Atomic mass of Cl = 35 g/mol

# moles of Cl = 2.21/35 = 0.0631 moles

Since moles of Ag << moles of Cl, silver is the limiting reagent.

Based on reaction stoichiometry: # moles of AgCl formed = 0.0155 moles

Enthalpy of formation of AgCl = 1.96 kJ/0.0155 moles = 126.5 kJ/mol

Ans: Formation enthalpy = 126.5 kJ/mol


6 0
3 years ago
Read 2 more answers
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Calcula el Tanta por ciento en peso de soluto
5 0
4 years ago
Assuming ideal solution behavior, what is the freezing point of a solution of 9.04 g of I2 in 75.5 g of benzene?
cluponka [151]

Answer:

3.1°C

Explanation:

Using freezing point depression expression:

ΔT = Kf×m×i

<em>Where ΔT is change in freezing point, Kf is freezing point depression constant (5.12°c×m⁻¹), m is molality of the solution and i is Van't Hoff factor constant (1 For I₂ because doesn't dissociate in benzene).</em>

Molality of 9.04g I₂ (Molar mass: 253.8g/mol) in 75.5g of benzene (0.0755kg) is:

9.04g ₓ (1mol / 253.8g) = 0.0356mol I₂ / 0.0755kg = 0.472m

Replacing in freezing point depression formula:

ΔT = 5.12°cm⁻¹×0.472m×1

ΔT = 2.4°C

As freezing point of benzene is 5.5°C, the new freezing point of the solution is:

5.5°C - 2.4°C =

<h3>3.1°C</h3>

<em />

3 0
3 years ago
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