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2 years ago

How many protons and electrons are contained in an atom of element 44?

Chemistry

2 answers:

vitfil [10]2 years ago

5 0

Answer:

Explanation:

protons and electrons have the same number:)

Andrews [41]2 years ago

5 0

Answer: element 44 = atomic # 44 44 protons 44 electrons.

Explanation:

You might be interested in

Given the following equation, what is the correct form of the conversion factor needed to convert the number of moles O2 to the

jek_recluse [69]

Answer:

Option A says we have 4 moles of Fe for each 3 moles O2

This is correct For 3 moles of O2 consumed, we need 4 moles of Fe to be reacted

Explanation:

Step 1: Data given

Step 2: The balanced equation

4Fe + 3O2 → 2Fe2O3

Step 3: Calculate the mol ratio

For 3 moles O2 we'll have 4 moles Fe

Option A says we have 4 moles of Fe for each 3 moles O2

This is correct For 3 moles of O2 consumed, we need 4 moles of Fe to be reacted

Option b says we have 2 mole Fe2O3 for each 4 moles Fe

This doesnt say anything about O2. So doesn't apply for this question.

Option C says we have 4 moles of Fe for each 2 moles Fe2O3

This is the same as option B, so doesn't apply for this question.

Option D says for each 3 moles of O2 we have 2 Fe2O3

This is true, but doesn't say anything about Fe so doesn't apply here.

4 0

2 years ago

Read 2 more answers

Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]$

For the given chemical reaction:

$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0

2 years ago

Calculate the volume in mL of a 0.708 M KOH solution containing 0.098 mol of solute

Olenka [21]

Molarity = mol/liter

0.708M = 0.098mol/L
Rearrange to find L:
0.098mol/0.708M = .138L

For every liter there is 1000 mL:
.138L • 1000mL =138mL KOH

7 0

3 years ago

Chemists investigated an unknown substance and found it to have the following characteristics:

bija089 [108]

The answer is option C.

That is it is a heterogeneous mixture.

Heterogeneous mixture have the following properties:

1. Different components could be observed in the substance.

2. Different samples of the substance appeared to have different proportions of the components.

3.The components could be easily separated using filters and sorting.

7 0

3 years ago

Read 2 more answers

Write and balance the half-reaction for the oxidation of white phosphorous P4 to the phosphate ion PO3^−4 in a basic solution.

aalyn [17]

Since the half-reaction is occurring in a basic solution, add 32OH− to each side of the equation to eliminate the H+ ions.

P₄ +16H₂O + 32OH⁻ ⟶ 4PO₃⁻⁴ + 32H⁺ +32OH⁻

Final reaction :

P₄ + 32OH⁻ ⟶ 4PO₃⁻⁴ + 16H₂O + 20e⁻

A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.

The concept of half-reactions is used to describe what occurs in an electrochemical cell, such as a Galvanic cell battery. Half-reactions can be written to describe both the metal undergoing oxidation (known as the anode) and the metal undergoing reduction (known as the cathode).

Half-reactions are often used as a method of balancing redox reactions. For oxidation-reduction reactions in acidic conditions, after balancing the atoms and oxidation numbers, one will need to add H+ ions to balance the hydrogen ions in the half reaction.

For oxidation-reduction reactions in basic conditions, after balancing the atoms and oxidation numbers, first treat it as an acidic solution and then add OH- ions to balance the H+ ions in the half reactions (which would give H2O).

Learn more about Half reactions here : brainly.com/question/2491738

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3 0

1 year ago