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3 years ago

Deﬁne atomic numberand mass number. Which can vary without changing the identity of the element?

Chemistry

1 answer:

Volgvan3 years ago

8 0

Explanation:

Atomic number is defined as the number of an element which represents total number of protons.

When an atom is neutral then it means it contains same number of protons and electrons.

On the other hand, atomic mass is defined as the sum of total number of protons and neutrons present in an atom.

Protons of every element remains fixed because it shows the identity of each element but if we change the number of neutrons then also identity of the atom will remain fixed. This is because changing the number of neutrons will not show any change in number of protons.

For example, $^{2}_{1}H$ and $^{3}_{1}H$ are isotopes of hydrogen and they have same number of protons but different number of neutrons.

Thus, we can conclude that number of neutrons can vary without changing the identity of the element.

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Instructions:Select the correct answer.

andriy [413]

Activity series of metals: K,Na,Mg,Al,Zn,Fe,Cu,Ag. Metals on the left are more reactive than metals on the right. For example Zn is more reactive than Fe and can displace him.
Reaction than can occur is: CuSO4(aq) + Fe(s) → FeSO4(aq) + Cu(s).

8 0

3 years ago

What is the identity of a 100. g sample of metal that, upon absorbing 4680 J of heat, increases in temperature by 52.0°C?

krok68 [10]

The metal is aluminium

Explanation:

Given:

Heat, q = 4680 J

Mass, m = 100g = 0.1kg

ΔT = 52°C

sample = ?

We know:

q = mcΔT

On substituting the value we get:

$4680 = 0.1 X c X(52)\\\\c = \frac{4680}{5.2} \\\\c = 900 J/Kg^oC$

Thus, the metal is aluminium which has a specific heat capacity of 900 J/kg°C

3 0

3 years ago

What is the volume of 3.00 mole of ideal gas at 100.0 C and 2.00 kPa

Novosadov [1.4K]

Answer:

The volume for the ideal gas is: 4647.5 Liters

Explanation:

Formula for the Ideal Gases Law must be applied to solve this question:

P . V = n . R . T

We convert the T° to K → 100°C + 273 = 373 K

We convert pressure value from kPa to atm.

2 kPa . 1atm/101.3 kPa = 0.0197 atm

We replace data in the formula.

V = ( n . R . T) / P → (3 mol . 0.082 . 373K) / 0.0197 atm =

The volume for the ideal gas is: 4647.5 Liters

8 0

3 years ago

What is the volume in liters of 321 g of a liquid with a density of 0.84 g/mL?

garri49 [273]

If we have 321 grams of a liquid, and the density is 0.84 g/mL, then we can easily find the volume of the liquid. We just need to take this 0.84 and multiply that by the number of grams. If we do 321 * 0.84, we get 269.64 mL. This is the volume that this liquid has.Remember this equation for future problems: V = D*M. V meaning volume, D meaning density, and M meaning mass. I hope this helps.

7 0

3 years ago

Read 2 more answers

For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l)

juin [17]

Answer:

The maximum amount of CO2 that can be formed is 9.15 grams CO2

O2 is the limiting reactant

There will remain 4.82 grams of benzene

Explanation:

Step 1: Data given

Mass of benzene = 7.53 grams

Mass of oxygen gas = 8.33 grams

Molar mass of benzene = 78.11 g/mol

Molar mass oxygen gas = 32.00 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles C6H6 = 7.53 grams / 78.11 g/mol

Moles C6H6 = 0.0964 moles

Moles O2 = 8.33 grams / 32.00 g/mol

Moles O2 = 0.2603 moles

Step 4: Calculate the limiting reactant

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles

There will remain 0.0964 - 0.0347 = 0.0617 moles benzene

This is 0.0617 moles * 78.11 g/mol = 4.82 grams benzene

Step 5: Calculate moles CO2

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.208 moles * 44.01 g/mol

Mass CO2 = 9.15 grams

4 0

3 years ago