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1 year ago

Using the activity series provided. Which reactants will form products?

2 answers:

5 0

The series of metals shown above is what we call the activity series. This is a guide for us to know what products are formed during a metal displacement reaction. A metal must be above the other metallic compound for displacement to occur.

1. Ag is way below the series compared to Na. No reaction occurs.
2. Al is above Fe, no reaction occurs.
3. Na is above Ni, no reaction will occur.
4. Fe is a more active metal than copper. A displacement reaction will occur.

Karo-lina-s [1.5K]1 year ago

4 0

The answer is 4: Fe+Cu(N03)2

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14) The central iodine atom in the ICl4- ion has __________ nonbonded electron pairs and __________ bonded electron pairs in its

masha68 [24]

Answer:

Two non bonded electron pairs and four bonded electron pairs

Explanation:

An image of the compound as obtained from chemlibretext is attached to this answer.

The ion ICl4- ion, is an AX4E2 ion. This implies that there are four bond pairs and two lone pairs of electrons. As expected, the shape of the ion is square planar since the lone pairs are found above and below the plane of the square. This is clear from the image attached.

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1 year ago

2.A solid block with a length of 6.0 cm, a width of 3.0 cm, and a height of 3.0 cm has a mass of 146 g. What is the block’s dens

posledela

2: <span>Volume V = a*b*c = 6.0*3.0*3.0 = 54.0 cm^3 density ρ = mass/volume = 146/54 = 2.70 g/cm^3
3: Volume = (27.8 -21.2) cm^3
mass = 22.4 g
density = 22.4/(27.8-21.2) g/cm^3

</span>

3 0

2 years ago

Read 2 more answers

N2 and H2 are mixed in 14:3 mass ratio. After certain time ammonia was found to be 40% by mol. The mole fraction of N2 at that t

son4ous [18]

Answer : The mole fraction of nitrogen will be 0.4615.

Explanation : When nitrogen ( $N_{2}$ )and hydrogen ( $H_{2}$ )are mixed, the mole ratio becomes 1 : 1.5,

Now we know that ( $H_{2}$ ) is acting as a limiting agent.

So at the time of when 0.4 moles of ( $NH_{3}$ ) is been formed it requires 0.4 moles of ( $N_{2}$ ) and 3.4 moles of ( $H_{2}$ )

So, we find the the remaining ( $N_{2}$ ) will be 0.6 and

( $H_{2}$ ) will be 0.3 mole present in mixture.

So, the mole fraction of ( $N_{2}$ ) becomes = 0.6 / (0.6 + 0.4 + 0.3) Which becomes = 0.4615

8 0

2 years ago

Read 2 more answers

A different student is given a 10.0g sample labeled CaBr2 that may contain an inert (nonreacting) impurity. Identify a quantity

grandymaker [24]

Answer:

Mass of Ca in sample, Mass of Br in sample, Number of moles of Ca in sample, Number of moles of Br in sample, Mass or moles of element other than Ca or Br in sample

Explanation:

The AP Classroom will not count your answer to this question as correct unless it includes at least one of the answers listed above. If you say that theanswer to this question is density, it will be marked as incorrect, I found that out the hard way when I used the answers that brainly gave me.

Good luck,

I applaud you for using the sources avalible to you, which is /definetly not/ cheeting.

8 0

1 year ago

Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. NH4l, CoBr3, Na2

seraphim [82]

Answer:

NaI > Na2SO4 > Co Br3

meaning that NaI has the highest freezing point, and Co Br3 has the lowest freezing point.

Explanation:

The freezing point depression is a colligative property.

That means that it depends on the number of solute particles dissolved.

The formula to calculate the freezing point depression of a solution of a non volatile solute is:

ΔTf = i * Kf * m

Where kf is a constant, m is the molality and i is the van't Hoff factor.

Molality, which is number of moles per kg of solvent, counts for the number of moles dissolved and the van't Hoff factor multipllies according for molecules that dissociate.

The higher the number of molecules that dissociate, the higher the van't Hoff, the greater the freezing point depression and the lower the freezing point.

As the question states that you assume equal concentrations (molality) and complete dissociation you just must find the number of ions generated by each solute, in this way:

NH4 I → NH4(+) + I(-) => 2 ions

Co Br3 → Co(+) + 3 Br(-) => 4 ions

Na2SO4 → 2Na(+) + SO4(2-) => 3 ions.

So, Co Br3 is the solute that generate more particles and that solution will exhibit the lowest freezing point among the options given, Na2SO4 is next and the NaI is the third. Ordering the freezing point from higher to lower the rank is NaI > Na2SO4 > CoBr3, which is the answer given.

4 0

1 year ago