All categories

3 years ago

Match the term to the correct definition or example.

Chemistry

1 answer:

GaryK [48]3 years ago

3 0

The amount of energy required to remove an electron from an atom. 2 ionization energy

A bond where electrons are shared unequally, resulting in a positively charged end and a negatively charged end of the molecule, 4 polar covalent bond

The tendency of an atom to attract a pair of electrons and bond with another atom; increases from bottom-left toward top-right of the periodic table. 3. Electronegativity

A bond where electrons are shared equally, or polar bonds in the molecule cancel each other out. 5. Non polar covalent bond

Shell or energy level where electrons are found. 1. Orbital

You might be interested in

Lakesha gave three tenths of her cookies to Bailey and five tenths of her cookies to Helen. What fraction of her cookies did Lak

Mashutka [201]

Answer:

a . eight tenths of her cookies

Explanation:

Let the total number of Lakesha's cookies be represented by x.

So that;

She gave three tenths to Bailey = $\frac{3}{10}$ of x

= $\frac{3x}{10}$

She gave five tenths to Helen = $\frac{5}{10}$ of x

= $\frac{5x}{10}$

Fraction of Lakesha's cookies given away = $\frac{3x}{10}$ + $\frac{5x}{10}$

= $\frac{3x+ 5x}{10}$

= $\frac{8x}{10}$

Thus, the fraction of cookies given away by Lakesha is $\frac{8}{10}$ .

6 0

3 years ago

How many of the planets have an orbital period of less than one Earth year?

GrogVix [38]

Since orbital period depends on how far you are from the sun, planets closer to the sun have a orbital period less than one earth year.

These planets are Mercury and Venus

8 0

3 years ago

Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equation

nordsb [41]

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

$C_2H_2(g) + \frac{5}{2}O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_{1} = -1,123 kJ$ ...[1]

$C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_{2} = -340 kJ$ ..[2]

$H_2(g) + \frac{1}{2}O_2(g)\rightarrow H_2O(l) ,\Delta H^o_{3} = -211 kJ$ ..[3]

$2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_{4} =?$ ..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

$\Delta H^o_{4}=2\times \Delta H^o_{2}+\Delta H^o_{3} - \Delta H^o_{1}$

$\Delta H^o_{4}=2\times (-340 kJ) + (-211 kJ) - (-1,123 kJ)$

$\Delta H^o_{4}=232 kJ$

The enthalpy for given reaction is 232 kilo Joules.

5 0

3 years ago

How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles

sattari [20]

1) Balance the chemical equation.

$3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2$

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4$

We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. This is the excess reactant.

3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2$

We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

$mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2$

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

$g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}$ $g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2$

The mass of Ca3(PO4)2 produced is 351 g Ca3(PO4)2.

Option D.

8 0

1 year ago

Balance the following skeleton reactions and identify the oxidizing and reducing agents:(b) P₄(s) → HPO₃²⁻(aq) + PH₃(g) [acidic]

m_a_m_a [10]

The balanced chemical equation is :

5P₄ + 36OH → 12HPO₃⁻² (aq) + 8PH₃ (acidic)

Here the oxidation number of P changed from 0 to -3 in PH₃ and increases from 0 to +3 in HPO₃⁻². When P₄ changes to PH₃ reduction reaction is taking place as there is addition of hydrogen and when P₄ changes to HPO₃⁻² oxidation takes place as there is addition of oxygen.

Thus clearly both reduction and oxidation are taking place.

Thus, we can infer that here P₄ is both oxidizing as well as reducing agent.

To know more about oxidation number here:

brainly.com/question/13182308

#SPJ4

5 0

2 years ago