How many moles of argon atoms are present in 11.2 L of argon gas at STP? Round your molar mass to whole numbers andgive your answer in the correct number of sig figs and units

Answer:

1.11 x 10²³ atoms O

Explanation:

To find the number of oxygen atoms in FeSO₄, you need to (1) convert FeSO₄ from grams to moles (using the molar mass), then (2) convert moles FeSO₄ to moles O (using the mole-to-mole ratio of FeSO₄), and then (3) convert moles O to atoms O (using Avogadro's Number). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs like the given number (7.00 = 3 sig figs).

Atomic Mass (Fe): 55.845 g/mol

Atomic Mass (S): 32.065 g/mol

Atomic Mass (O): 15.999 g/mol

Molar Mass (FeSO₄): 55.845 g/mol + 32.065 g/mol + 4(15.999 g/mol)

Molar Mass (FeSO₄): 151.906 g/mol

1 mole FeSO₄: 1 mole Fe, 2 mole S, 4 moles O

Avogadro's Number:

6.022 x 10²³ atoms = 1 mole

 7.00 g FeSO₄             1 mole                 4 moles O            6.022 x 10²³ atoms
-----------------------  x  -------------------  x  -----------------------  x  ----------------------------
                                   151.906 g           1 mole FeSO₄                  1 mole

=  1.11 x 10²³ atoms O

10.81amu

In order to get the average atomic mass of an element, we need the following parameters:

• Natural Abundance (NA),: The percentage of atoms for an element that is a specific isotope.

• Mass (m) ,of each isotope

For the given element (Boron-10 and Boron-11), the natural abundances are 19.78% and 80.22% respectively.

The atomic masses of Boron-10 and Boron-11 are 10.0129amu and 11.0093amu respectively

The formula for calculating the average atomic mass of the element is expressed as:

Substitute the given parameters into the formula to have:

Simplify the resulting expression to have:

Therefore the average atomic mass of Boron is 10.81amu to two decimal places.

Answer:

Explanation:

Here, we want to get the amount of heat released in KJ

From the change in enthalpy given and the equation of reaction, we know that 2 moles of HBr would lead to that amount of heat

Now, let us get the actual amount of heat released

We need to get the actual number of moles of HBr produced

Mathematically, we can calculate that by dividing the mass of HBr by its molar mass

We have that as:

From the reaction information:

-72 KJ was released by 2 moles

x KJ would be released by 0.0122 mol

To get the value of x, we have it that:

By definition, the percent composition of an atom in a compound is its mass percentage in the formula.

That is, if we have 1 mol of NaHCO₃, we have also 1 mol of H (because there is only on H for each molecule).

So, we calculate the mass of this 1 mol of NaHCO₃ and the mass of 1 mol of H and calculate the percentage.

In equations, we want the following:

Since these are ratios, we doesn't matter if we talk about 1, 2 or any number of moles, but 1 mol is easier because the molecular and atomi masses are for 1 mol.

The molecular mass of NaHCO₃ is:

Which means that we have approximately 84.006 grams of NaHCO₃ in 1 mol of it.

The atomic mass of H is:

Which means that we have approximately 1.008 grams of H in 1 mol of it.

Now, we can take the percentage of mass of H:

So, the percentage composition of H in NaHCO₃ is approximately 1.20%.

Answer:

CH3CH2NH2 is a weak base.

Explanation:

CH3CH2NH2 is a weak base since its Kb is small, and thus it partially dissociates.

Answer

The element that is oxidized is Si and the one that is reduced is F₂.

Explanation

Si + 2F₂ → SiF₄

The given reaction is an oxidation-reduction (redox) reaction:

Oxidation: Si → Si⁴⁺ + 4e⁻

Reduction: 2F₂ + 4e⁻ → 4F⁻

Si is a reducing agent, and F₂ is an oxidizing agent.

An oxidizing agent gains electrons and is reduced in a chemical reaction.

A reducing agent loses electrons and is oxidized in a chemical reaction.

Therefore, the element that is oxidized is Si and the one that is reduced is F₂.

Ideal gas law is valid only for ideal gas not for vanderwaal gas. Therefore the carbon dioxide in the cartridge inflate a 3.05 L  mountain bike tire to 104.3psi pressure.

Ideal gas equation is the mathematical expression that relates pressure volume and temperature.

Mathematically,

PV=nRT

where,

P = pressure

V= volume=3.05 L

n =number of moles=1mole(assumed as it is not given in question)

T =temperature =  302 K

R = Gas constant = 0.0821 L.atm/K.mol

P × 3.05 L  =1 mole× 0.0821 L.atm/K.mol ×  302 K

P =8.12atm

1 atm = 14.7 psi

Hence Pressure in psi  = 8.12atm×14.7 psi = 119.49 psi

Pressure by the gas= Total pressure - Atmospheric pressure = 119.49 - 14.7 psi = 104.3psi

Therefore the carbon dioxide in the cartridge inflate a 3.79 L mountain bike tire to 104.3psi pressure.

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The theoretical percentage yield of copper  is 70%

This is further explained below.

Generally, the equation for  yield is mathematically given as

[tex]yield =\frac{\text { mote no. of product }}{\text { mote no. of reactant }} \times 100$[/tex]

Here, Based on Theoretical Yield

[tex]\begin{aligned} \text { mole rio. of reactant } &=-\frac{0.5}{63.546} \text { mole } \\ \end{aligned}$[/tex]

Based on the Actual yield

mole rio. of Product=0.35/63.546 mole

(Here atomic man of Cu=63.546g/mol )

Therefore, we apply the initially stated equation for %yield

So,  %yield =[tex]\frac{\frac{0.35}{63.546}}{\frac{0.5}{(63.546}} \times 100 \%$[/tex]

=[tex]\frac{0.35}{0.5} \times 100 \%\\[/tex]

=0.7 *100%

=70%

In conclusion,  the percentage yield is 70%

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Answer

B. 1.7

Explanation

As a rule, an electronegativity difference of 2 or more on the Pauling scale between atoms leads to the formation of an ionic bond. A difference of less than 2 between atoms leads to covalent bond formation.

Therefore, the largest possible electronegativity difference for a bond to be covalent is 1.7

1) First let's write the equation. It is a combustion reaction, so:

C₃H₈ + O₂ ---> CO₂ + H₂O

and balance the equation (same number of atoms of each element on both sides of the equation):

C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O

Reactant side:

C - 3

H - 8

O - 10

Product side:

C - 3

O - 10

H - 8

2) Now let's transform 85 grams of CO₂ into mole. For this, we use the following equation:

mole = mass/molar mass

Molar mass of CO₂ is: (1×12) + (2×16) = 44 g/mol

mole = 85/44

mole = 1.9 mol of CO₂

3) Now we use the proportion of the balanced equation:

1 mol of C₃H₈ ---- 3 mol of CO₂

x mol of C₃H₈ ----- 1.9 mol of CO₂

x = 0.6 mol of C₃H₈

4) Now we transform mole of C₃H₈ into grams using its molar mass.

molar mass of C₃H₈ is: (3×12) + (8×1) = 44 g/mol

mass = mole × molar mass

mass = 0.6 × 44

mass of C₃H₈ = 28 g

Answer: a) mass of C₃H₈ = 28 g

For alternative b we follow the same process starting from step 3:

3)Now we use the proportion of the balanced equation:

5 mol of O₂ ---- 3 mol of CO₂

x mol of O₂ ----- 1.9 mol of CO₂

x = 3.16 mol of O₂

4) Now we transform mole of O₂ into grams using its molar mass.

molar mass of O₂ is: (2×16) = 32 g/mol

mass = mole × molar mass

mass = 3.16 × 32

mass of O₂ = 101 g

Answer: b) mass of O₂ = 101 g

For alternative c we follow the same process starting from step 3:

3) Now we use the proportion of the balanced equation:

4 mol of H₂O ---- 3 mol of CO₂

x mol of H₂O ----- 1.9 mol of CO₂

x = 0.84 mol of H₂O

4) Now we transform mole of H₂O into grams using its molar mass.

molar mass of H₂O is: (2×1) + (1×16) = 18 g/mol

mass = mole × molar mass

mass = 0.84 × 18

mass of H₂O = 15 g

Answer: c) mass of H₂O = 15 g

The relative atomic mass of the given element is 40.372 amu.

The relative atomic mass of an element is considered as the sum of the isotopes masses each multiplied by the percentage which is found in nature.

The formula which is used to calculate the relative atomic mass is

Relative atomic mass = sum of all atomic masses of isotopes × fractional abundance

Given,

Mass of isotopes 1 = 19.19 amu

Mass of isotopes 2 = 20.99 amu

Mass of isotopes 3 = 21.99 amu

Fractional abundance of isotope 1 = 0.9092

Fractional abundance of isotope 2 = 0.0025

Fractional abundance of isotope 3 = 0.0883

By substituting all the values, we get

[( 19.19 × 0.9092) + (20.99 × 0.0025) + (21.99 × 0.883)]

= 17.447 + 0.052 + 22.873

= 40.372 amu.

Thus, we concluded that the relative atomic mass of the given element is 40.372 amu.

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We have the following formulas for the given compounds:

1. Magnesium nitride ---> Mg3N2

2. Lithium oxide ---> Li2O

3. Aluminum sulfite ---> Al2(SO4)3

4. Copper (II) bicarbonate ---> Cu(HCO3)2

5. Sodium nitrate ---> NaNO3

For the given formulas we have the following names:

6. SrI2 ---> Strontium iodide

7. Ba3(PO4)2 --->Barium phosphate

8. (NH4)2O ---> Ammonium oxide

9. Fe(ClO)3 ---> Iron(III) hypochlorite

10. ZnCrO4 ---> zinc chromate

Answer

Refrigeration slows the rate at which food is being spoiled

Explanation

One of the importance of storing foods at cold temperatures (refrigeration) is to slow the growth of microorganisms, thereby limiting food poisoning while preserving food's nutritional qualities and good taste.

Therefore, the impact the cargo ship refrigeration systems have on the banana industry is:

Refrigeration slows the rate at which food is being spoiled

Answer:

Mixtures can be classified on the basis of particle size into three different types: solutions, suspensions and colloids. The components of a mixture retain their own physical properties.

Sorry for the bad English, love from Vanuatu!

Explanation:

Answer

The percent yield = 83.36%

Explanation

Given:

Experimental yield = actual yield = 15.68 g

Theoretical yield = 18.81 g

What to find:

The percent yield for the reaction.

Step-by-step solution:

The percent yield for the reaction can be calculated using the formula below:

Hence, the percent yield for the reaction is 83.36%

Answer

Explanation

Given:

Mass of Fe2O3 = 80 g

Equation:

What to find:

The grams of H2 required to completely convert 80g of Fe2O3.

Step-by-step solution:

From the equation of reaction;

3 moles of H2 completely react with 1 mole of Fe2O3

Note: Molar mass of H2 is 2.016 grams per mole and Molar mass of Fe2O3 is 159.69 g/mol

This implies; (3 x 2.016 g) = 6.048 grams H2 completely react with 159.69 grams Fe2O3.

Therefore, x grams H2 will completely convert 80 grams Fe2O3.

Cross multiply and divide both sides by 159.69 grams Fe2O3.

x grams H2 is now equal to

Therefore the grams of H2 required to completely convert 80g of Fe2O3 is 3.0 grams

As the question gave us the formula in which we have to use to calculate the volume of this type of wood:

V = m/d

We have:

m = 1700 grams

d = 0.85

Now we add these values into the formula:

V = 1700/0.85

V = 2000 cm3, letter D

Empirical formula:

Step 1

Information already provided

The mass percent composition:

40.9 % C

4.58 % H

54.5 % O

Information needed: from the periodic table

For C) 1 mol = 12.01 g

For H) 1 mol = 1.008 g

For O) 1 mol = 15.99 g

---------------------------

Step 2

A sample of 100 g is assumed, so:

40.9 % C => 40.9 g C

4.58 % H => 4.58 g H

54.5 % O => 54.5 g O

--------------------------

Step 3

Convert mass into moles:

40.9 g C x (1 mol/12.01 g) = 3.40 moles C

4.58 g H x (1 mol/1.008 g) = 4.54 moles H

54.5 g O x (1 mol/15.99 g) = 3.40 moles O

------------------------

Step 4

All moles calculated in step 3 need to be divided by the smallest one.

3.40 moles C/3.40 moles = 1

4.54 moles H/3.40 moles = 1.33

3.40 moles O/3.40 moles = 1

-----------------------

Step 5

Integer numbers are needed, so let's multiply by 3 all of them in step 4

Therefore,

For C) 3

For H) 3.99 = 4 approx.

For O) 3

All these numbers calculated will be the subindexes in ascorbic acid

Answer:

Empirical formula: C3H4O3

Metals normally have different densities, so we can try to determine which metal is this by its density.

Assuming it is pure, the density of the metal is its mass divided by its volume:

Now, we need to look for some table with densities of various metals and see which one has the density we found, 11.4 g/cm³.

Since we don't have one, we can look for one. In it, we can see that the only metal with this density is lead.

So, the metal in the question should be lead.

Answer:

Explanation:

The new pressure can be found by using the formula

P1V1 = P2V2

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we're finding the new pressure P2 we make P2 the subject

We have

[tex]p_2 = \frac{p1v1}{v2} \\ [/tex]

P1 = 1.7 atm

V1 = 0.6 L

V2 = 0.9 L

We have

[tex]p_2 = \frac{1.7 \times 0.6}{0.9} = 1.13333...\\ [/tex]

We have the final answer as

Hope this helps you

The following lists consists of ionic compounds except carbon disulfide (CS₂).

Barium hydroxide , Ba(OH)₂ is an ionic compound.

zinc carbonate, ZnCO₃ is an ionic compound.

ammonium sulfate , (NH₄)₂SO₄  is an ionic compound.

calcium chloride, CaCl₂  is an ionic compound.

carbon disulfide, CS₂ is not an ionic compound. In carbon disulfide both the elements are non metallic elements. The bond formed between atoms are by sharing of electron known as covalent bond due to very little difference in electronegativity.

magnesium nitrate, Mg(NO₃)₂  is an ionic compound.

sodium sulfate, Na₂SO₄  is an ionic compound.

copper(II) oxide, CuO  is an ionic compound.

potassium nitride KNO₃  is an ionic compound.

aluminium sulfide, Al₂S₃  is an ionic compound.

sodium sulfite, Na₂S  is an ionic compound.

calcium fluoride, CaF₂  is an ionic compound.

Thus, The following lists consists of ionic compounds except carbon disulfide (CS₂).

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Mole equation: [tex]n=\dfrac{m}{MM}[/tex]

To find mass given number of atoms:

We're given:

In the chemical formula, which is Ag, there is only 1 atom.

Divide [tex]1.81*10^{23}[/tex] atoms by 1 atom to get the number of molecules in the silver ring:

[tex]1.81*10^{23}\div 1\\=1.81*10^{23}[/tex]

Therefore, there are [tex]1.81*10^{23}[/tex] molecules in the silver ring.

Now, divide [tex]1.81*10^{23}[/tex] molecules by Avogadro's number to find n:

[tex]\dfrac{1.81*10^{23}}{6.02*10^{23}}\\\\=\dfrac{1.81}{6.02}\\\\=\dfrac{1.81}{6.02}\\\\= 0.30066[/tex]

Therefore, the sample has 0.30066 mol.

Finally, solve for the mass using the moles equation:

[tex]n=\dfrac{m}{MM}[/tex]

⇒ Rearrange the equation:

[tex]m=n*MM[/tex]

⇒ MM of Ag = 107.87 g/mol

⇒ Plug in given information:

[tex]m=0.30066* 107.87\\m=32.4[/tex]

Therefore, the mass of the ring is 32.4 g.

32.4 g

KOH. Option B is correct

A chemical compound are made up of more than one element combined together. According to the question, we need to determine the formula that represents a compound.

The compound there is KOH since it contains three elements (Potassium, Oxygen and Hydrogen)

The [OH⁻] of the solution with pH of 8.13 is 1.35 * 10-6 M

pH is the measurement of the acidity or basicity of a compound by measuring [H⁻] ions in the solution. It ranges from 0 to 14 with acidic range from 0 – 7 and basic range from 7-14.

pOH is the measurement of the acidity or basicity of a compound by measuring [OH⁻] ions in the solution.

Thus, pH + pOH = 14

pOH = 14 - pH = 14 – 8.13 = 5.87

pOH = - log ([OH⁻])

- log ([OH⁻]) = 5.87

Log ([OH⁻]) = - 5.87

[OH⁻] = 10 ^ - 5.87   = 0.00000134896

[OH⁻] = 1.35 * 10⁻⁶ M

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Lithium nitride is an ionic compound and the chemical formula of lithium nitride is Li₃N and the charges on the ions is Li⁺¹N³⁻ 3 valence electron in lithium and 5 valence electron in nitrogen

Ionic compound is the ion compound these ions are atom that gain or lose electron resulting in a net positive and negative charges and lithium nitride is an ionic compound and in that Li⁺¹N³⁻ is the ion and the arrangement of the valency (outer shell) electrons are 3 valence electron in lithium and 5 valence electron in nitrogen

X = 3 electron

O = 7 electron

And that's why lithium nitride has 10 electron

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A component that many space agencies use is liquid hydrogen, which can be abbreviated as LH2.

And another fuel very used in rockets is kerosene which is a hydrocarbons mixture of other compounds that fuel.

Answer:

Most liquid chemical rockets use two separate propellants: a fuel and an oxidizer. Typical fuels include kerosene, alcohol, hydrazine and its derivatives, and liquid hydrogen. Many others have been tested and used. Oxidizers include nitric acid, nitrogen tetroxide, liquid oxygen, and liquid fluorine.

Explanation:

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mL and cm³ have a 1 to 1 conversion, so we have:

But we don't have this option, so we will need to find another.

The other are in m³, so we can use the conversion from cm to m to get this:

So, we can apply this to what we have:

We have an option with 0.054m³, so the correc alternative is 0.054 m³.