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2 years ago

Problem

Chemistry

1 answer:

postnew [5]2 years ago

4 0

Answer:

C) Potassium, K, and calcium, Ca

Explanation:

Given pair of elements:

Ni and Cu

Potassium and Ca

Nitrogen and oxygen

Solution:

We know that number of protons and number of electrons are always equal and number of electron or number of proton is called atomic number.

Number of protons given elements:

Ni number of protons = 28

Copper number of protons = 29

Combine number of protons = 57

K number of protons = 19

Ca number of protons = 20

Combine number of protons = 39

Ni number of protons = 7

O number of protons = 8

Combine number of protons = 15

Thus, option C is correct.

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What is "carbon neutral"

kati45 [8]

its is making no net release of carbon dioxide to the atmosphere,especially through offsetting emissions by planting trees

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2 years ago

Read 2 more answers

How many molecules OF2 would have a mass of .132 g

BlackZzzverrR [31]

<h3>Answer:</h3>

1.47 × 10²¹ molecules OF₂

<h3>General Formulas and Concepts: </h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation: </h3>

<u>Step 1: Define</u>

0.132 g OF₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of O - 16.00 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of OF₂ - 16.00 + 2(19.00) = 54.00 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 0.132 \ g \ OF_2(\frac{1 \ mol \ OF_2}{54.00 \ g \ OF_2})(\frac{6.022 \cdot 10^{23} \ molecules \ OF_2}{1 \ mol \ OF_2})$
Divide/Multiply: $\displaystyle 1.47204 \cdot 10^{21} \ molecules \ OF_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.47204 × 10²¹ molecules OF₂ ≈ 1.47 × 10²¹ molecules OF₂

3 0

2 years ago

Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0

2 years ago

Is this right?If not what did I do wrong ?

Eddi Din [679]

Hey there,

It seems we do not have an image. Mind supplying us with one?

6 0

2 years ago

Which is the most likely to be reduced?

NISA [10]

Answer : The correct option is, $Zn^{2+}$

Explanation :

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. That means, the loss of electrons takes place.

Or we can say that, oxidation reaction occurs when a reactant losses electrons in the reaction.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. That means, the gain of electrons takes place.

Or we can say that, reduction reaction occurs when a reactant gains electrons in the reaction.

According to the electrochemical series, $Zn^{2+}$ most likely to be reduced because

Hence, the ion most likely to be reduced is $Zn^{2+}$ .

3 0

2 years ago