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Alona [7]
3 years ago
10

What weighs more, a gram of of iron or a gram of salt?

Chemistry
1 answer:
USPshnik [31]3 years ago
4 0

Answer:

none of them.......

You might be interested in
How many unpaired electrons are present in the ground state of an atom from each of the following groups?
Paha777 [63]

Explanation:

General electronic configuration for the given groups are as follows.

For 4A(14) : ns^{2}np^{4}

This means that there are 4 unpaired electrons in group 4A(14).

For 7A(17) : ns^{2}np^{5}

So, there are 7 unpaired electrons present in group 7A(17).

For 1A(1) : ns^{1}

Hence, there is only one unpaired electron present in group 1A(1).

For 6A(16) : ns^{2}np^{4}

So, there are 6 unpaired electrons present in group 6A(16).

3 0
4 years ago
Calculate the freezing point and boiling point of a solution containing 14.2 gg of naphthalene (C10H8)(C10H8) in 111.0 mLmL of b
krek1111 [17]

Answer:

The freezing point of solution = -0.34 °C

The boiling point of solution = 82.98 °C

Explanation:

Step 1: Data given

Mass of naphthalene = 14.2 grams

Molar mass naphthalne = 128.17 g/mol

Volume of benzene = 111.0 mL

Density of benzene = 0.877 g/mL

Kf(benzene)=5.12°C/m

Freezing point benzene = 5.5 °C

Kb(benzene)=2.53°C/m

Boiling point benzene = 80.1 °C

Step 2: Calculate mass of benzene

Mass benzene = density * volume

Mass benzene = 0.877 g/mL * 111.0 mL

Mass benzene = 97.3 grams

Step 3: Calculate moles naphthalene

Moles naphthalene = mass naphthalene / molar mass napthalene

Moles napthalene = 14.2 grams / 128.17 g/mol

Moles naphthalene = 0.111 moles

Step 4: Calculate molality

Molality = moles naphthalene / mass benzene

Molality = 0.111 moles / 0.0973 kg

Molality = 1.14 molal

Step 5: Calculate freezing point  of a solution

ΔT = i*kf*m

ΔT = 1 * 5.12 °C/m * 1.14 m

ΔT = 5.84 °C

ΔT = T(pure solvent) − T(solution)

The freezing point of solution = T pure -  ΔT

5.5 - 5.84 = -0.34 °C

The freezing point is -0.34 °C

Step 6: Calculate boiling point  of a solution

ΔT = i*kb*m

ΔT = 1 * 2.53 °C/m * 1.14 m

ΔT = 2.88 °C

ΔT = Tb (solution) - Tb (pure solvent)

The boiling point of solution = T pure +  ΔT

The boiling point of solution = 80.1 °C + 2.88

The boiling point of solution = 82.98 °C

6 0
3 years ago
9. A bond formed between a silicon atom and an oxygen atom is likely to be
Elanso [62]
<h2><u><em>Answer: </em></u></h2>

<u><em>The answer is </em></u>

<u><em>O ionic </em></u>

<h2><u><em>Explanation: </em></u></h2>

  • <em>A bond formed between a silicon atom and an oxygen atom is likely to be: ionic</em>
5 0
3 years ago
Identify the number of core and valence electrons for each atom.
AlexFokin [52]

Answer:

1. Core electrons = 1s²; Valence electrons = 2s² and 2p³

2. Core electrons = 1s², 2s²,m2p⁶; Valence electrons = 3s² and 3p⁵

Explanation:

The electrons in an atom of an element are generally divided into two groups: the core electrons and the Valence electrons

Valence electrons are those electrons which are located of or found in the outermost shell or highest energy level of an atom. These valence electrons are the one that determine the chemical reactivity of an atom as they can participate in bond formation between atoms of the same element or with atoms of other elements. The valence electrons also, generally determine the group to which an element will belong to.

Core electrons on the other hand, are those electrons which are found within the innermost shell or lowest energy levels of the atom. They are strongly attracted to the nucleus of atom and do not take part in chemical bonding. However, they play the important role of shieldingnthe valence electrons from the positive charge of the nucleus thereby assisting the valence electrons in bond formation.

3 0
3 years ago
When a 3.80 g sample of C8H18(l) is burned in a bomb calorimeter, the temperature of the calorimeter rises by 27.3 oC. The heat
Fudgin [204]

<u>Answer:</u> The enthalpy of the reaction is -5112.5 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

q=c\Delta T

where,

q = heat absorbed

c = heat capacity of calorimeter = 6.18 kJ/°C

\Delta T = change in temperature = 27.3°C

Putting values in above equation, we get:

q=6.18kJ/^oC\times 27.3^oC=168.714kJ

Heat absorbed by the calorimeter will be equal to the heat released by the reaction.

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of octane = 3.80 g

Molar mass of octane = 114 g/mol

Putting values in above equation, we get:

\text{Moles of octane}=\frac{3.80g}{114g/mol}=0.033mol

To calculate the enthalpy change of the reaction, we use the equation:

\Delta E=\frac{q}{n}

where,

q = amount of heat released = -168.714 kJ

n = number of moles = 0.033 moles

\Delta E = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta E=\frac{-168.714kJ}{0.033mol}=-5112.5kJ/mol

Hence, the enthalpy of the reaction is -5112.5 kJ/mol

4 0
3 years ago
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