We need to prepare 0.0021 M solution of C2SO4. But there are only 80.7 grams of the chemical available. What is the maximum volume that can be prepared?

Answers

Answer 1

Answer:

320 L

Explanation:

We'll begin by calculating the number of mole in 80.7 g of C₂SO₄. This can be obtained as follow:

Mass of C₂SO₄ = 80.7 g

Molar mass of C₂SO₄ = (12×2) + 32 + (16×4)

= 24 + 32 + 64

= 120 g/mol

Mole of C₂SO₄ =?

Mole = mass /molar mass

Mole of C₂SO₄ = 80.7 / 120

Mole of C₂SO₄ = 0.6725 mole

Finally, we shall determine the volume. This can be obtained as follow:

Mole of C₂SO₄ = 0.6725 mole

Molarity = 0.0021 M

Volume =?

Molarity = mole / Volume

0.0021 = 0.6725 / Volume

Cross multiply

0.0021 × Volume = 0.6725

Divide both side by 0.0021

Volume = 0.6725 / 0.0021

Volume ≈ 320 L

Thus, the volume that can be prepared is approximately 320 L.


Related Questions

what is the difference between a chemical bonds formed in the molecules of 02 and the chemical bonds formed in crystals of a NaCI

Answers

Answer:

O2 is a covalent substance while NaCl is an ionic substance

Explanation:

In O2 molecule, the bond is between 2 oxygen atoms which are non - metals. Thus, this is a covalent bond since it involves 2 non metals.

Whereas, for the NaCl molecule, the bond is between a metal sodium (Na) and a non metal Chloride(Cl) and thus we can say this is an ionic bond.

Thus the difference is that O2 is a covalent substance while NaCl is an ionic substance.

Using a balanced chemical equation, and 2.50 g of sodium hydrogen carbonate as the reactant,
what is the expected (theoretical) yield of sodium carbonate (grams)? The Formula Weight (FW) of
sodium hydrogen carbonate is 84.01 g and sodium carbonate is 105.99 g.

Answers

Answer:

1.58 g

Explanation:

Step 1: Write the balanced equation

2 NaHCO₃ ⇒ Na₂CO₃ + H₂O + CO₂

Step 2: Calculate the moles corresponding to 2.50 g of NaHCO₃

The molar mass of NaHCO₃ is 84.01 g/mol.

2.50 g × 1 mol/84.01 g = 0.0298 mol

Step 3: Calculate the moles of Na₂CO₃ produced

The molar ratio of NaHCO₃ to Na₂CO₃ is 2:1. The moles of Na₂CO₃ produced are 1/2 × 0.0298 mol = 0.0149 mol

Step 4: Calculate the mass corresponding to 0.0149 moles of Na₂CO₃

The molar mass of Na₂CO₃ is 105.99 g/mol.

0.0149 mol × 105.99 g/mol = 1.58 g

Why are prefixes not needed in naming ionic compounds?

Answers

Answer:

when naming ionic compounds — those are only used in naming covalent molecular compounds. Do NOT use prefixes to indicate how many of each element is present; this information is implied in the name of the compound. since iron can form more than one charge. Ionic Compounds Containing a Metal and a Polyatomic Ion.

When 250. mg of eugenol, the molecular compound responsible for the odor of oil of cloves, was added to 100. g of camphor, it lowered the freezing point of camphor by 0.62 8C. Calculate the molar mass of eugenol.

Answers

Answer:

Molar mass for eugenol is 161.3 g/mol

Explanation:

This question talks about freezing point depression:

Our solute is eugenol.

Our solvent is camphor.

Formula to state the freezing point depression difference is:

ΔT = Kf . m . i where

ΔT = Freezing T° of pure solvent - Freezing T° of solution

In this case ΔT = 0.62°C

Kf for camphor is: 37°C /m

As eugenol is an organic compund, i = 1. No ions are formed.

To state the molar mass, we need m (molal)

Molal are the moles of solute in 1kg of solvent. Let's replace data:

0.62°C = 40 °C/m . m . 1

0.62°C / 40 m/°C = 0.0155 m

We convert mass of camphor from g to kg = 100 g . 1kg / 1000g = 0.1 kg

0.0155 molal = moles of solute / 0.1 kg

0.0155 m/kg . 0.1 kg = 0.00155 moles

We know that these moles are contained in 250 mg, so the molar mass will be:

0.25 g / 0.00155 mol = 161.3 g/mol

Notice, we convert mg to g, for the units!

Consider the reaction C4H10O + NaBr + H2SO4 → C4H9Br + NaHSO4 + H2O. If 45.0 g of C4H10O reacts with 67.1 g of NaBr and 97.0 g of H2SO4to yield 60.0 g of C4H9Br, calculate the percent yield of the reaction.

Answers

Answer:

Percent yield  = 72.07 %

Explanation:

Our reaction is:

C₄H₁₀O + NaBr + H₂SO₄ → C₄H₉Br + NaHSO₄ + H₂O

It is correctly balanced.

Let's determine which is the limiting reagent:

45 g . 1 mol / 74 g = 0.608 moles of C₄H₁₀O

67.1 g . 1 mol / 102.9 g = 0.652 moles of NaBr

97 g . 1 mol / 98 g = 0.990 moles of sulfuric acid

Ratio is always 1:1, so for 1 mol of NaBr and 1 mol of sulfuric acid we need 1 mol of C₄H₁₀O. We have 0.652 moles of NaBr, we need the same amount of C₄H₁₀O and we have 0.990 moles of acid, we need the same amount of C₄H₁₀O; we only have 0.608 moles, that's why C₄H₁₀O is the limiting reactant, there's no enough C₄H₁₀O.

Ratio is also 1:1, between reactant and product.

1 mol of C₄H₁₀O produces 1 mol of C₄H₉Br

Then, 0.608 moles will produce 0.608 moles of C₄H₉Br

We convert moles to mass: 0.608 mol . 136.9 g/mol = 83.25 g

That's the 100 % yield reaction

Percent yield  = (Yield produced / Theoretical yield) . 100

Percent yield = (60 g / 83.25 g) . 100 = 72.07 %

In an experiment 25.0 mL of 0.100 M KI was diluted to 50.0 mL. Calculate the molarity of the diluted solution

Answers

Answer:

The molarity is "0.050 M".

Explanation:

The given values are:

M1 = 0.100 M

M2 = ?

V1 = 25.0 mL

V2 = 50.0 mL

As we know,

⇒ [tex]M1\times V1=M2\times V2[/tex]

Or,

⇒ [tex]M2=\frac{M1\times V1}{V2}[/tex]

By putting the values, we get

          [tex]=\frac{0.100\times 25}{50}[/tex]

          [tex]=\frac{2.5}{50}[/tex]

          [tex]=0.05 \ M[/tex]

An enzyme increases the reaction rate by:________. a. stabilizing of transition state. b. shifting the reaction equilibrium. c. increasing the probability of product formation. d. All of the answers are correct. e. None of the answers is correct.

Answers

Enzymes provide activation energy for the reactions they catalyze.

Classify each molecule as an alcohol, ketone, or aldehyde based on its name. Propanone (acetone) Choose... Ethanal Choose... 3-phenyl-2-propenal Choose... Butanone Choose... Ethanol Choose... 2-propanol Choose...

Answers

Answer:

1.) Propanone (ketone)

2.) Ethanal( aldehyde)

3.) 3-phenyl-2-propenal (aldehyde)

4.) Butanone (ketone)

5.) Ethanol ( alcohol)

6.) 2-propanol (alcohol)

Explanation:

In organic chemistry, ALCOHOL ( also known as alkanol) are compounds in which hydroxyl groups are linked to alkyl groups. They can be considered as being derived from the corresponding alkanes by replacing the hydrogen atoms with hydroxyl groups. The hydroxyl group is the functional group of the alcohol as it is responsible for their characteristic chemical properties. A typical example of alcohol is ethanol and 2-propanol.

Alkanals or ALDEHYDES have the general formula RCHO while alkanones or KETONES have the general formula RR'CO where R and R' may be alkyl or aryl groups. The main similarity between these two classes of compounds is the presence of the carbonyl group. In aldehydes, there is a hydrogen atom attached to the carbon In the carbonyl group while there is none on the ketones.

Some common examples of ketones are Propanone, Butanone while examples of aldehydes are Ethanal and 3-phenyl-2-propenal

(c) m X is an ion in which group of the periodic table is the element from which X is formed?​

Answers

Explanation:

Iron has 2 atoms and 3atoms.

So,X=2,3

2. How many joules of heat are released when 32g of water cools down from 71%
specific heat of water is 4.184 J/gºC)
How many kilojoules is this?

Answers

he says he doesnt know sorry

complete the following steps.
Remember to follow lower numbered rules first.
Na2CO3(aq) + Pb(OH)2(aq) → NaOH (?) + PbCO3(?)
a. Write a balanced chemical equation. (1 pt)
b. If a reaction occurs, write the balanced
chemical equation with the proper states of matter
(i.e. solid, liquid, aqueous) filled in. If no reaction
occurs, write “No reaction.” (1 pt)
c. If a reaction occurs, write the net ionic equation
for the reaction. If no reaction occurs, write "no
reaction.” (1 pt)

Answers

Answer:

See explanation

Explanation:

a) The balanced reaction equation is;

Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH + PbCO3

b) When we include states of matter;

Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH(aq) + PbCO3 (s)

c) Complete ionic equation;

2Na^+(aq) + CO3^2-(aq) + Pb^2+(aq) + 2OH^-(aq) ----> 2Na^+(aq) + 2OH^-(aq) + PbCO3(s)

Net Ionic equation;

Pb^2+(aq) + CO3^2-(aq) ----> PbCO3(s)

Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. (a) AgCl(s) in 0.025 M NaCl (b) CaF2(s) in 0.00133 M KF (c) Ag2SO4(s) in 0.500 L of a solution containing 19.50 g of K2SO4 (d) Zn(OH)2(s) in a solution buffered at a pH of 11.45\

Answers

Answer:

Explanation:

a) AgCl(s) in 0.025 M NaCl

Equation:  AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

Initial conc :    S            O               O

equili conc :    O            S                S

                  NaCl(s) ⇒ Na⁺ (aq) + Cl⁻ (aq)

Initial conc :  0.025      0           0

equili conc :     0          0.025    0.025

Therefore the concentration:  Ag⁺ = 6.4 * 10^-9 M,  Cl⁻  = 0.025 M

attached below is the detailed solution of the

Help!!!!!!!!!
I'm using plato

Answers

Answer:

- Two black balls: they represent a diatomic molecule composed by two atoms of the same element.

- One black ball and two black balls: they represent a compound formed by two different elements.

- One gray ball and two black balls: they represent a compound formed by two different elements.

- Two black-dotted balls: they represent a diatomic molecule composed by two atoms of the same element.

Explanation:

Hey there!

In this case, according to the given information, we can firstly bear to mind the fact that each ball color represents a different element, for that reason we can tell the following:

- Two black balls: they represent a diatomic molecule composed by two atoms of the same element.

- One black ball and two black balls: they represent a compound formed by two different elements.

- One gray ball and two black balls: they represent a compound formed by two different elements.

- Two black-dotted balls: they represent a diatomic molecule composed by two atoms of the same element.

Regards!

79.1,3-Butadiene molecule contains how many sigma and pi bond

Answers

Explanation:

3 sigma and 3 pieee

okok kkk

A container is filled to a volume of 55.2 L at 61 °C. While keeping the
temperature constant, the volume is reduced to 28.8 L and the pressure at
the end was recorded to be 8.53 atm. What was the initial pressure inside
the container, in units of atm?

Answers

Answer:

4.45 atm

Explanation:

Applying,

PV = P'V'............ Equation 1

Where P = Initial pressure of the container, V = Initial volume of the container, P' = Final pressure of the container, V' = Final volume of the container.

make P the subject of the equation

P = P'V'/V........... Equation 2

From the question,

Given: V = 55.2 L, P' = 8.53 atm, V' = 28.8 L

Substitute these values into equation 2

P = (8.53×28.8)/55.2

P = 4.45 atm

Answer:

[tex]\boxed {\boxed {\sf 4.45 \ atmospheres}}[/tex]

Explanation:

We are asked to find the pressure given a change in volume. The temperature remains constant, so we are only concerned with volume and pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula for this law is:

[tex]P_1 V_1= P_2V_2[/tex]

The initial pressure is unknown, but the volume starts at 55.2 liters.

[tex]P_1 * 55.2 \ L = P_2V_2[/tex]

The volume is reduced to 28.8 liters and the pressure is 8.53 atmospheres.

[tex]P_1 * 55.2 \ L = 8.53 \ atm * 28.8 \ L[/tex]

We are solving for the initial pressure, so we must isolate the variable P₁. It is being multiplied by 55.2 liters. The inverse operation of multiplication is division, so we divide both sides of the equation by 55.2 L.

[tex]\frac {P_1 * 55.2 \ L }{55.2 \ L}= \frac{8.53 \ atm * 28.8 \ L}{55.2 \ L}[/tex]

[tex]P_1= \frac{8.53 \ atm * 28.8 \ L}{55.2 \ L}[/tex]

The units of liters (L) cancel.

[tex]P_1= \frac{8.53 \ atm * 28.8 }{55.2}[/tex]

[tex]P_1=\frac{245.664 }{55.2 } \ atm[/tex]

[tex]P_1 = 4.45043478261 \ atm[/tex]

The original measurements of volume and pressure have 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredths place. The 0 in the thousandths place tells us to leave the 5.

[tex]P_1 \approx 4.45 \ atm[/tex]

The initial pressure inside the container is approximately 4.45 atmospheres.

Solid potassium chlorate (KClO3)(KClO3) decomposes into potassium chloride and oxygen gas when heated. How many moles of oxygen form when 48.1 gg completely decomposes

Answers

Answer:

0.59 mol O₂

Explanation:

The balanced chemical equation for the decomposition of potassium chlorate (KClO₃) to produce potassium chloride (KCl) and oxygen gas (O₂) is the following:

2 KClO₃ → 2 KCl + 3 O₂

According to the equation, 3 moles of O₂ are produced from 2 moles of KClO ⇒ conversion factor: 3 mol O₂/2 mol KClO₃

Now, we calculate the number of moles of KClO₃ there is in 48.1 g, by dividing the mass into the molecular weight (Mw) of O₂:

Mw(KClO₃) = 39.1 g/mol + 35.4 g/mol + (16 g/mol x 3) = 122.5 g/mol

moles KClO₃ = mass KClO₃/Mw(KClO₃) = 48.1 g/(122.5 g/mol) = 0.3926 mol KClO₃

Finally, we multiply the moles of KClO₃ by the conversion factor to calculate the moles of O₂ produced:

0.3926 mol KClO₃ x 3 mol O₂/2 mol KClO₃ = 0.59 mol O₂

A hypothetical A-B alloy of composition 53 wt% B-47 wt% A at some temperature is found to consist of mass fractions of 0.5 for both and phases. If the composition of the phase is 92 wt% B-8 wt% A, what is the composition of the phase

Answers

Answer:

the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Explanation:

Given the data in the question;

Co = 53 or [ 53 wt% B-47 wt% A ]

W∝ = 0.5 = Wβ

Cβ = 92 or [ 92 wt% B-8 wt% A ]

Now, lets set up the Lever rule for W∝ as follows;

W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]

so we substitute our given values into the expression;

0.5 = [ 92 - 53 ] / [ 92 - C∝ ]

0.5 = 39 /  [ 92 - C∝ ]

0.5[ 92 - C∝ ] = 39

46 - 0.5C∝  = 39

0.5C∝ = 46 - 39

0.5C∝ = 7

C∝ = 7 / 0.5

C∝ = 14  or [ 14 wt% B-86 wt% A ]

Therefore, the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

If H2O acts as an acid in a reaction, what would be its conjugate base?

Answers

Answer:

since H2O is an acid, by the Arrhenius definition, it would donate a proton. Thus, the conjugate base is OH~

How many atoms are present in 0.45 moles of P4010

Answers

Answer:

80g

Explanation:

mass oxygen present in 1 mole of p4010

16×10=160gm

similarly

for 0.5 moles of p4010 160/2= 80gm

The number of atoms present in 0.45 moles of P₄O₁₀ is 1.08 x 10²³ atoms.

To determine the number of atoms, we use Avogadro's number, which states that there are approximately 6.022 x 10²³ particles (atoms, molecules, or formula units) in one mole of a substance.

In this case, we are given 0.45 moles of P₄O₁₀. To calculate the number of atoms, we multiply the number of moles by Avogadro's number:

Number of atoms = 0.45 moles P₄O₁₀ x (6.022 x 10²³ atoms / 1 mole)

Number of atoms = 2.7139 x 10²³ atoms

Rounding to three significant figures, the number of atoms present in 0.45 moles of P₄O₁₀ is approximately 1.08 x 10²³ atoms.

To learn more about atoms   here

https://brainly.com/question/3127831

#SPJ2

Gaseous BF3 and BCl3 are mixed in equal molar amounts. All B-F bonds have about the same bond enthalpy, as do all B-Cl bonds. Compare the numbers of microstates to explain why the mixture tends to react to form BF2Cl(g) and BCl2F(g

Answers

Solution :

[tex]$BF_3 (g) + BCl_3 (g) \rightarrow BF_2 Cl + BCl_F(g)$[/tex]

Explanation 1 :

Spontaneity of the reaction is based on two factors :

-- the tendency to acquire a state of minimum energy

-- the energy of a system to acquire a maximum randomness.

Now, since there isn't much difference in the bond enthalpies of B-F and B-Cl. So, we can say the major driving factor is tendency to acquire a state of maximum randomness.

Explanation 2 :

A system containing the [tex]\text{"chemically mixed"}[/tex] B halides has a [tex]\text{greater entropy}[/tex] than a system of [tex]$BCl_3$[/tex] and [tex]BF_3[/tex].

It has the same number of [tex]\text{gas phase molecules}[/tex], but more distinguishable kinds of [tex]\text{molecules}[/tex], hence, more microstates and higher entropy.

Calculate the molarity of a 17.5% (by mass) aqueous solution of nitric acid. Select one: a. 2.74 m b. 4.33 m c. 0.274 m d. 3.04 m e. The density of the solution is needed to solve the problem.

Answers

Answer:

Option e.

Explanation:

Molarity is the concentration that indicates moles of solute in 1 L of solution.

We have another concentration, percent by mass.

Percent by mass indicates mass of solute in 100 g of solution.

Our solute is HNO₃, our solvent is water.

17.5 g of nitric acid is the mass of solute. We can convert them to moles:

17.5 g . 1mol / 63g = 0.278 moles

We do not have volume of solution. We assume the mass is 100 g because the percent by mass but we need density to state the volume.

Density = Mass / Volume

Mass / Density = Volume

Once we have the volume, we need to be sure the units is in L, to determine molarity

M = mol /L

Predict the reactants of this chemical reaction.

products: KClO4 + H2O
reactants: ?

Answers

Answer:

KOH + HClO4 = K(ClO4) + H2O

Explanation:

KClO4 + H2O : mix and match the letters if you need to guess

Classify each of the four compounds as a conjugated, isolated, or cumulated diene. Compound A: Two alkenes are joined by a sigma bond. Compound A is a: cumulated diene conjugated diene isolated diene Compound B: Two alkenes are joined by a C H 2 group. Compound B is : isolated diene conjugated diene cumulated diene Compound C: Two alkenes are joined by C H 2 C H 2. Compound C is a: conjugated diene isolated diene cumulated diene Compound D: A cyclohexene has a double bond between carbons 1 and 2. Carbon 3 is an s p 2 carbon that is bonded to another s p 2 carbon with an alkyl substituent. Compound D is a: isolated diene conjugated diene cumulated diene

Answers

Explanation:

Conjugated diene is the one that contains alternate double bonds in its structure. That means both the double bonds are separated by a single bond.

Cumulated diene is the one that contains two double bonds on a single atom. This means it has two double bonds continuously.

Isolated double-bonded compound has a single bond isolated by two to three single bonds.

Compound A: Two alkenes are joined by a sigma bond.

For example:

[tex]-CH_2=CH-CH=CH2-[/tex]

It is a conjugated diene.

Compound B: Two alkenes are joined by a C H 2 group.

It is a cumulative diene.

Compound C: Two alkenes are joined by C H 2 C H 2.

Then it is an isolated alkene.

Compound D:  A cyclohexene has a double bond between carbons 1 and 2. Carbon 3 is an sp 2 carbon that is bonded to another s p 2 carbon with an alkyl substituent.

Hence, compound D is a conjugated diene.

The density of mercury is 13.6 g/cm3, What is its density in mg/mm3?

Answers

Answer:

Density of mercury is 13600 kg

Which of the following is the best definition of a physical change?

A. Ice melting into water.
B. A change that occurs without changing the identity of the substance.
C. Something that can be observed or measured while changing the identity of the substance.
D. A nail rusting.

Answers

You’re best choice is B

When we test sucrose with seliwanoff's test what would the result be positive or negative ? Before and after hydrolysis of sucrose .

Answers

I need more explain?

Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work!

Convert 3.00 x 10^21 atoms of copper to moles.

Convert 2.25 x 10^18 molecules of carbon dioxide to moles.

Answers

Answer:

1) 0.00498 mol Cu.

2) 0.00000374 mol CO₂

Explanation:

Question 1)

We want to convert 3.00 * 10²¹ copper atoms into moles. Note that 3.00 is three significant figures.

Recall that by definition, one mole of a substance has exactly 6.022 * 10²³ amount of that substance. In other words, we have the ratio:

[tex]\displaystyle \frac{1\text{ mol}}{6.022\times 10^{23} \text{ Cu}}[/tex]

We are given 3.00 * 10²¹ Cu. To cancel out the Cu, we can multiply it by our above ratio with Cu in the denominator. Hence:

[tex]\displaystyle 3.00 \times 10^{21} \text{ Cu} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} \text{ Cu}}[/tex]

Cancel like terms:

[tex]=\displaystyle 3\times 10^{21} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} }[/tex]

Simplify:

[tex]\displaystyle = \frac{3\text{ mol Cu}}{6.022 \times 10^{2}}[/tex]

Use a calculator:

[tex]= 0.004981... \text{ mol Cu}[/tex]

Since the resulting answer must have three significant figures:

[tex]= 0.00498\text{ mol Cu}[/tex]

So, 3.00 * 10²¹ copper atoms is equivalent to approximately 0.00498 moles of copper.

Question 2)

We want to convert 2.25 * 10¹⁸ molecules of carbon dioxide into moles. Note that 2.25 is three significant digits.

By definition, there will be 6.022 * 10²³ carbon dioxide molecules in one mole of carbon dioxide. Hence:

[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ CO$_2$}}{1\text{ mol CO$_2$}}[/tex]

To cancel the carbon dioxide from 2.25 * 10¹⁸, we can multiply it by the above ratio with the carbon dioxide in the denominator. Hence:

[tex]\displaystyle 2.25\times 10^{18} \text{ CO$_2$} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23} \text{ CO$_2$}}[/tex]

Cancel like terms:

[tex]\displaystyle= 2.25\times 10^{18} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23}}[/tex]

Simplify:

[tex]\displaystyle = \frac{2.25 \text{ mol CO$_2$}}{6.022\times 10^5}}[/tex]

Use a calculator:

[tex]=0.000003736...\text{ mol CO$_2$}[/tex]

Since the resulting answer must have three significant figures:

[tex]= 0.00000374\text{ mol CO$_2$}[/tex]

So, 2.25 * 10¹⁸ molecules of carbon dioxide is equivalent to approximately 0.00000374 moles of carbon dioxide.

Answer:

Explanation:

by definition, 1 mole contains 6.02 x 10^23 of atoms (for elements) or molecules (for compounds)

3.00 x 10^21 atoms of copper / 6.02 x 10^23 of atoms

= 0.004983 moles of copper

= 4.98 x 10^(-3) moles of copper

2.25 x 10^18 molecules of carbon dioxide / 6.02 x 10^23 of molecules

= 0.000003737 moles of carbon dioxide

= 3.74 x 10^(-6) moles of carbon dioxide

Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 417 mL .

Answers

Answer:

m = 1125.9 g.

Explanation:

Hey there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the definition of density as mass divided by volume:

[tex]d=\frac{m}{V}[/tex]

Thus, we solve for the mass in the equation to obtain:

[tex]m=d*V[/tex]

Then, we plug in the values to obtain:

[tex]m=2.70g/mL*417mL\\\\m=1125.9g[/tex]

Regards!

According to the kinetic theory, all matter is made of moving particles, which measurement of matter is directly proportional to the
average kinetic energy of the particles?

Answers

Answer: Kelvin temperature of a substance

What type of bond does hafnium oxide have?
A. ionic
B. covalent
C. Metallic

Answers

Answer:

C) METALLIC IS THE CORRECT ANSWER

Explanation: I just did the exam

Other Questions
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