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3 years ago

How many atoms, in PCI3, and how can I tell

Chemistry

1 answer:

Leya [2.2K]3 years ago

8 0

Answer:

one Phosphorus atom and three Chlorine atoms, having a chemical formula of PCl3. It is a volatile liquid that reacts with water and releases HCl gas. It is a toxic compound but is used in several industries.Jul 23, 2021

Explanation:

i am so srry if this is wrong

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The answer is 197
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3 years ago

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Question 1 (True/False Worth 2 points)

pochemuha

1) Answer is: False.

Balanced chemical reaction: Pb + AlCl₃ → no reaction.

The reactivity series is a series of metals from highest to lowest reactivity. Metal higher in the reactivity series will displace another.

Aluminium (Al) is higher in activity series than lead (Pb).

That means aluminium (Al) is stronger reducing agent than lead and gives electrons easier.

This series are used to summarize information about the reactions of metals with acids or water, double displacement reactions (more reactive metals displace metals with lower reactivity) and the extraction of metals from their ores.

2) Answer is: Na3PO4 + 3KOH → 3NaOH + K3PO4, because K retains the same charge throughout the reaction.

This chemical reaction is double displacement reaction - cations (K⁺ and Na⁺) and anions (PO₄³⁻⁻ and OH⁻) of the two reactants switch places and form two new compounds.

Na₃PO₄ is sodium phosphate.

KOH is potassium hydroxide.

NaOH is sodium hydroxide.

K₃PO₄ is potassium phosphate.

3) Answer is: Zn + H2SO4 → ZnSO4 + H2.

Single replacement is reaction where neutral element metal or nonmetal become an ion as it replaces another ion in a compound.

In this balanced chemical reaction zinc (Zn) replaces hydrogen (H).

Zinc changes oxidation number from 0 to +2 (this is oxidation) and hydrogen change oxidation number from +1 to 0 (reduction).

4) Answer is: A single replacement reaction takes place because sodium is more reactive than hydrogen.

Balanced chemical reaction:

2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g).

Single replacement is reaction where neutral element metal or nonmetal become an ion as it replaces another ion in a compound.

In this balanced chemical reaction sodium (Na) replaces hydrogen (H).

Sodium changes oxidation number from 0 to +1 (oxidation) and hydrogen change oxidation number from +1 to 0 (reduction).

5) Answer is: True.

For example, balanced chemical reaction:

Ca + 2HCl → H₂ + CaCl₂.

Reduction hald reaction: 2H⁺ + 2e⁻→ H₂.

Oxidation half reaction: Ca⁰ → Ca²⁺ + 2e⁻.

In this balanced chemical reaction calcium (Ca) replaces hydrogen (H).

In this chemical reaction, calcium lost two electrons (oxidation, change oxidation number from 0 to +2) and hydrogen gain that two electrons (reduction, change oxidation number from +1 to 0).

6) Answer is: It is a single replacement reaction, and the cations in the two ionic compounds are different.

For example, balanced chemical reaction:

Ba + MgSO₄ → BaSO₄(s) + Mg.

In this balanced chemical reaction barium (Ba) replaces magnesium (Mg), because barium is higher at activity series.

Sulfate anion (SO₄²⁻) is the same in two ionic compounds, but cations (Mg²⁺ and Ba²⁺) changes.

Single Displacement is reaction where neutral element metal or nonmetal become an ion as it replaces another ion in a compound.

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3 years ago

PLEASE HELP ME ASAP!!!! PLEASE!!!!!!!!!! I will give brainliest and 20 points!! PLEASE!

Vika [28.1K]

#1
The rate of diffusion of a gas is inversely proportional to the square root of its molecular weight.
#2
shape, molecular weight. I don't know for sure though
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O2, N2, Ar, H2O vapor

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3 years ago

Excess oxygen gas (O2) reacts with 244g of Iron (Fe) to produce 332 g of Fe2O3. What is the percent yield?

Andru [333]

Answer:

95.15%

Explanation:

To calculate the percent yield, we need the following formula:

$\% yield=\dfrac{actual yield}{theoreticalyield}\times100\%$

Solving for the theoretical yied, you need to predict how much of the product will be produced if we USE up the given.

Our given is 224g of Fe and we to get the theoretical yield, we need to figure out how much product will Fe produce supposing that we use up all the reactant.

First thing we do is get the balance equation of this chemical reaction:

4Fe + 3O₂ → 2Fe₂O₃

We get the ratio between Fe and the the product, Fe₂O₃

$\dfrac{4moles of Fe}{2molesofFe_{2}O_{3}}=\dfrac{2moles of Fe}{1moleofFe_{2}O_{3}}$

This basically means that we need 2 moles of Fe to produce 1 mole of Fe₂O₃. We'll use this later.

Now we let's use our given:

We need to first convert our given to moles. To do this, we need to determine how many grams there are of the reactant for every mole. We need to first get the atomic mass of the elements involved in the substance:

Iron(1)

Fe = 55.845(1) = 55.845g/mole(1)

Then we use this to convert grams to moles

$244g\times\dfrac{1mole}{55.845g}=4.369moles$

This means that there are 4.396moles of Fe in 244g of Fe.

This we will use to see how many moles of product we can produce given the moles of reactant by using the reactant:rproduct ratio.

$4.369moles of Fe\times\dfrac{1moleofFe_{2}O_{3}}{2molesofFe}=2.185 moles of Fe_{2}O_{3}$

So given 4.693 moles of Fe we can produce 2.185 moles of Fe₂O₃

The next step is to get how many grams of product there are given our calculation. We do this again by getting how many grams of Fe₂O₃ there are in 1 mole.

Fe(2) O(3)

Fe₂O₃=55.845(2) + 15.999(3)

= 111.69 + 47.997 =159.687g/mol

We then use this to solve for how many grams of product there are in 2.185 moles.

$2.185moles\times\dfrac{159.687g}{1mole}=348.92g$

This is our theoretical yield 348.92g of Fe₂O₃.

We can finally use our percent yield equation. Our actual yield is given by the probelm, 332g of Fe₂O₃ and we solved for our theoretical yield which is 348.92g of Fe₂O₃. We plug this in our formula and solve.

$\%yield=\dfrac{actual yield}{theoreticalyield}\times100\%$

$\%yield=\dfrac{332gofFe_{2}O_{3}}{348gofFe_{2}O_{3}}\times100\%=95.15\%$

So the answer is 95.15%

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3 years ago

Is Cytoplasm found in animal cells,plant cells or in both?

nataly862011 [7]

Answer:

cytoplasm is found in both plant and animal cell.

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