Kevin's supervisor, Jill, has asked for an update on today's sales, Jill is pretty busy moving back and forth between different store locations. How can Kevin most effectively deliver an update to her ? a) Call with a quick update Ob ) Send a detailed text message c ) Book a one-hour meeting for tomorrow morning d) Send a detailed email

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)

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]

he says he doesnt know sorry

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

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

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

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 %

Answer:

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

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

3 sigma and 3 pieee

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I need more explain?

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

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.

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.

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

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Answer:

Density of mercury is 13600 kg

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 ]

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

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!

Explanation:

Iron has 2 atoms and 3atoms.

So,X=2,3

Answer:

209

83 Bi

Explanation:

213 213 - 4 4

85 At = 85 - 2 Y + 2 He

Given question is:

Given a food with a relative humidity of 81%, what is the water activity in that food?

Explanation:

Water activity in food can be determined by using the below-shown formula:

[tex]water activity=\frac{equilibrium relative humidity}{100}[/tex]

Equilibrium is established between the vapor pressure of food and the surrounding air media.

Thus, for the given food the relative humidity is 81%.

hence, its water activity is

[tex]81/100\\=0.81[/tex]

Answer:

gravity, ground, friction, rolling resistance, and air resistance.

Answer:

Mass of C = 47.37g

Mass of H = 10.59g

Mass of O = 42.04g

The total mass of these elements is 100g, taking a proportion of their molar masses.

C = 47.37/12= 3.95

H = 10.59/1 = 10.59

O = 42.04/16= 2.63.

Dividing through with the smallest proportion which is 2.63

C=3.95/2.63 = 1.5

H =10.59/2.63 =4

O = 2.63/2.63= 1

Multiplying through by 2 to get a whole number.

C = 1.5x2 = 3

H= 4x2 = 8

O = 1x2= 2

The empirical formula is C3H6O2

(Empirical formula)n= molecular mass

(C3H8O2)n =228.276

(12x3 +8+16x2)n= 228.276

76n = 228.276

n = 228.276/76

n = 3

Molecular formula = Empirical formula

=(C3H8O2)3 = C9H24O6

The molecular formula is C9H24O6

Answer:

% yield of reaction is 26.4

Explanation:

The reaction is:

Mg + I₂ →  MgI₂

Our reactants are magnessium and iodine. We determine the moles of each to find the limiting reactant:

10 g . 1mol / 24.3 g = 0.411 moles of Mg

95 g . 1mol / 253.8g = 0.374 moles of I₂

Ratio is 1:1. For 1 mol of Mg we need 1 mol of iodine

For 0.411 moles, we need the same amount, but we only have 0.374 moles of iodine, that's why the gas is the limiting reactant.

As ratio is 1:1 again, 0.374 moles of iodine can produce 0.374 moles of MgI₂

We determine the mas (theoretical yield): 0.374 mol . 278.1 g/mol = 104 g

To calculate the percent yield:

% yield = (yield produced /theoretical yield) . 100

% yield = (27.5 g/ 104g) . 100 = 26.4

[tex] \boxed{\boxed{\mathfrak{ 1\: mole \:of \:Na_2Cr_2O_2\: = \:it's \:Gram\: Mol. \: mass}} }[/tex]

[tex]\underline{ \mathfrak{ Gram \:molecular \:mass \:of \: \red{ Na_2Cr_2O_2}}}[/tex]

= 2 × 23 + 2 × 52 + 2 × 16

= 182 grams

1 mole of [tex]Na_2Cr_2O_2[/tex] weighs = 182 g

8 moles weigh = 8× 182

=[tex] \mathfrak{\blue {\boxed{\underline {1456 \: grams}}}} [/tex]

or

[tex] \mathfrak{\blue {\boxed{\underline {1. 46 \:kg }}}} [/tex]

Answer:

[tex]\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}[/tex]

Explanation:

A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).

Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.

Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.

Answer:

hydrogen bonding

Explanation:

just took the test :D