Esters are formed from the reaction of an alcohol and a carboxylic acid. Identify the alcohol and carboxylic acid combination necessary to make each pictured ester.

Answer:

B) Eating a antacid to calm your upset stomach.

Explanation:

Answer:

A. 1.02 moles .

Explanation:

Hello!

In this case, given the ideal gas equation, as we need to solve for moles, we divide both sides by RT to get:

[tex]n=\frac{PV}{RT}[/tex]

Thus, by plugging in the pressure and temperature at STP (1.00 atm and 273.15 K respectively) we obtain:

[tex]n=\frac{1.00atm*22.9L}{0.08206\frac{atm*L}{mol*K}*273.15K}\\\\n=1.02mol[/tex]

Therefore, the correct answer is A. 1.02 moles

Best regards!

Answer:

The answer is 79.904.

Explanation:

We assume you are converting between grams Br and mole.

You can view more details on each measurement unit:

molecular weight of Br or mol

The SI base unit for amount of substance is the mole.

1 grams Br is equal to 0.012515018021626 mole.

Note that rounding errors may occur, so always check the results.

Use this page to learn how to convert between grams Br and mole.

Type in your own numbers in the form to convert the units!

give 5 stars rn

Which of the statements about liquids and gases are true?

The particles in liquids and gases can move past one another freely.

Which of the statements about liquids and gases are true?

Both liquids and gases have a definite volume.

The particles in liquids and gases can move past one another freely

Both liquids and gases have a definite shape.

Neither liquids nor gases have a definite shape.

The particles in liquids and gases can move past one another freely.

Pounds of force pressing on a rectangle : F = 173.86 lbs

Given

P at sea level = 14.7 lbs/in²

Area of rectangle : A = 76.3 cm²

Required

Pounds of force

Solution

P = F/A

F= P.A

Conversion :

1 cm² = 0.155 in²

76.3 cm² = 11.827 in²

Input the value :

F = 14.7 lbs/in² x 11.827 in²

F = 173.86 lbs

Answer:

7x10⁴.

Explanation:

Hello!

In this case, when rounding big number to a representation with less significant figures than it, we need to make sure we remove the spare figures and round up whether the next digit is five or more.

In such a way, for 74,000, we notice that 4 is not enough to round up the 7, that is why the number would contain a 7 only; moreover, to get rid of the thousand places, we need to introduce an exponent in 4 places, which means that the result would be:

7x10⁴.

Best regards!

Answer:

XY4Z2- dsp2

XY4Z- sp3d

XY5Z- sp3d2

XY2Z3- sp3d

XY2Z- sp2

XY3Z- sp3

XY2Z2- sp3

XY3Z2- sp3d

XY2 - sp

,XY3- sp2

XY4- sp3

,XY5- sp3d

XY6- sp3d2

Explanation:

In the answer box I have shown the hybridization patterns of the various arrangement of bond pairs and lone pairs.

Hybridization was put forward to explain the shapes of molecules under the valence bond theory. All molecules having the same hybridization pattern are based on the same geometry. Deviations from the expected geometry may arise due to the presence of lone pairs.

Answer:

E = 7.11 × [tex]10^{-7}[/tex]  Cal

Explanation:

given data

efficiency = 25 %

solution

first we get here E by the power that is express as

p = [tex]\frac{E}{t}[/tex]     ...........1

E = p × t

E = 1 ×  2 × [tex]\frac{1hp}{746w} \times \frac{1hr}{3600 s}[/tex]  

E = 7.447 × [tex]10^{-7}[/tex] J

and

Ein is get by efficiency

Ein = [tex]\frac{Eout}{\eta }[/tex]        ..................2

Ein = [tex]\frac{7.447 \times 10^{-7}}{.25 }[/tex]

Ein = 29.788 × [tex]10^{-7}[/tex]  J

so required energy is

E = 29.788 × [tex]10^{-7}[/tex]  ×  [tex]\frac{1Cal}{4.184J}[/tex]

E = 7.11 × [tex]10^{-7}[/tex]  Cal

16.7 g H₂O

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Stoichiometry

Step 1: Define

[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

[Given] 1.85 mol NaOH

Step 2: Identify Conversions

[RxN] 2 mol NaOH → 1 mol H₂O

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Stoichiometry

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

16.6685 g H₂O ≈ 16.7 g H₂O

Answer: [tex]CsF[/tex] > [tex]SrF_2[/tex] >  [tex]OF_2[/tex]  > [tex]F_2[/tex]

Explanation:

The percent ionic character is used to determine if a bond is more ionic or covalent. More is the percent ionic character, the more ionic the molecule is. If the electronegativity difference between two elements is higher than 1.7 it will be considered as ionic.

The electronegativity difference for CsF is (3.98-0.79) = 3.19

The electronegativity difference for [tex]OF_2[/tex] is = (3.98-3.44) = 0.54

The electronegativity difference for [tex]SrF_2[/tex] is (3.98-0.95) = 3.0

The electronegativity difference for [tex]F_2[/tex] is  0.

Thus decreasing order of  ionic character of the bonds is CsF >  [tex]SrF_2[/tex] >  [tex]OF_2[/tex]  > [tex]F_2[/tex]

Answer:

Chalcogen

Explanation:

Answer:

Upwelling! Its the process where the deep cold water rises to the surface! :)

Explanation:

Hope this helps! :)

Answer:

upwelling

Explanation:

Answer:

Therefore, the specific heat capacity of the iron is 0.567J/g.°C.

Note: The question is incomplete. The complete question is given as follows:

A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm.

Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits

Explanation:

Using the formula of heat, Q = mc∆T  

where Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat capacity (J/g∙°C), ∆T = change in temperature (°C)

When the hot iron is placed in the water, the temperature of the iron and water attains equilibrium when the temperature stops changing at 27.6 °C. Since it is assumed that heat exchange occurs only between the iron metal and water; Heat lost by Iron = Heat gained by water

mass of iron  = 59.1 g, c = ?, Tinitial = 85.0 °C, Tfinal = 27.6 °C

∆T = 85.0 °C - 27.6 °C = 57.4 °C

mass of water = 100.0 g, c = 4.184 J/g∙°C, Tinitial = 23.0 °C, Tfinal = 27.6 °C

∆T = 27.6°C - 23.0°C = 4.6 °C

Substituting the values above in the equation; Heat lost by Iron = Heat gained by water

59.1 g * c * 57.4 °C  = 100.0 g * 4.184 J/g.°C * 4.6 °C

c = 0.567 J/g.°C

Therefore, the specific heat capacity of the iron is 0.567 J/g.°C.

Answer: a) [tex]1.7\times 10^{-4}[/tex]

b) [tex]3.4\times 10^{-4}[/tex]

Explanation:

The reaction is :

[tex]2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)[/tex]

Rate = Rate of disappearance of [tex]N_2O_5[/tex] = Rate of appearance of [tex]NO_2[/tex]

Rate =  [tex]-\frac{d[N_2O_5]}{2dt}[/tex] = [tex]\frac{d[NO_2]}{4dt}[/tex]

Rate of disappearance of [tex]N_2O_5[/tex] = [tex]\frac{\text {change in concentration}}{time}[/tex] = [tex]\frac{0.100-0.066}{200-0}=1.7\times 10^{-4}[/tex]

a) Rate of disappearance of [tex]N_2O_5[/tex] = [tex]-\frac{d[N_2O_5]}{2dt}[/tex]

Rate of appearance of [tex]NO_2[/tex] = [tex]\frac{d[NO_2]}{4dt}[/tex]

b) Rate of appearance of [tex]NO_2[/tex] =  [tex]\frac{d[NO_2]}{dt}=2\times 1.7\times 10^{-4}}=3.4\times 10^{-4}[/tex]

A) Find the rate of disappearance of [tex]N_2O_5[/tex] from t = 0 s to t = 200s

[tex]Rate = \frac{1}{2}(\frac{-\delta N_2O_5}{\delta t})\\\\Rate = -\frac{1}{2}(\frac{0.066 - 0.100}{200 - 0})\\\\Rate = 8.5*10^{-5}[/tex]

B) Find the rate of appearance of [tex]NO_2[/tex] from t = 0 s to t = 200s

According to rate law,

[tex]\frac{1}{2}(\frac{-\delta N_2O_5}{\delta t}) = \frac{1}{4}(\frac{\delta NO_2}{\delta t})\\\\8.5*10^{-5} = \frac{1}{4}(\frac{\delta NO_2}{\delta t})\\\\\frac{\delta NO_2}{\delta t} = 4 * 8.5*10^{-5}\\\\Rate = 3.4*10^{-4}[/tex]

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Answer:

1. 10moles

2. Find the summarized equation below

Explanation:

Mole = mass ÷ molar mass

Molar mass of oleum (H2S2O7) = 1(2) + 32(2) + 16(7)

= 2 + 64 + 112

= 178g/mol

According to the information provided, the mass of oleum is 1780g, hence, the number of moles in it is:

Mole = 1780/178

Mole = 10moles

2. The summarized equation of the contact process is:

1) S + O2 → SO2

2) 2SO2 + O2 → 2SO3

3) SO3 + H2SO4 (conc. sulfuric acid) → H2S2O7

4) H2S2O7 + H2O → 2H2SO4 (dilute sulfuric acid)

Answer:

D

Explanation:

I hope this helps and have a great day

Answer:

I think the answer is A the Afracan plate

Explanation:

hope this helps:)

Answer: Coefficient 0

subscript for V 2

subscript for O 5

Explanation: yes

There is no coefficient in V₂O₅ and the subscript of vanadium V is 2 and the subscript of oxygen is 5.

Vanadium pentoxide is an ionic compound formed by donating electrons from the vanadium metal to oxygen.  The chemical formula of vanadium pentoxide is V₂O₅.

The chemical formula of a compound is written in terms of the chemical symbol of each constituent elements and the number of their atoms. The number of atoms is written as subscript for the chemical symbols.

In V₂O₅, there are are 2 vanadium atoms and 5 oxygen atoms. Coefficients are numbers prior to the formula in reaction. Here there is no coefficient and the subscript for V is 2 and that of O is 5.

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Answer:

A yan hehehehe

salamat puso ma ah

Explanation:

Let the mass of the compound be 100g.

There are 60.2g of sulfur.

Moles of sulfur

= 60.2g / (32.07g/mol) = 1.877mol

There are 100g - 60.2g

= 39.8g of manganese.

Moles of manganese

= 39.8g / (54.94g/mol) = 0.724mol

Mole ratio of manganese to sulfur

= 0.724mol : 1.877mol = 1 : 2.5.

Mass (in g) of the excess reactant (P₄) left = 7.626 g

Given

Reaction

P4(s)+6Cl2(g)→4PCl3(l)

45.57 g P4 and 130.2 g Cl2

Required

mass (in g) of the excess reactant left

Solution

mol P₄ (MW=124 g/mol) :

45.57 : 124 = 0.3675

mol Cl₂(MW=71 g/mol) :

130.2 : 71 = 1.834

mol : coefficient of reactant : P₄ : Cl₂ :

= 0.3675/1 : 1.834/6

= 0.3675 : 0.306

Cl₂ as limiting reactant(smaller ratio)

Reacted mol of P₄ (as an excess reactant):

=1/6 x 1.834

= 0.306

Unreacted mol of P₄ :

= 0.3675 - 0.306

= 0.0615

Mass of P₄(left) :

= 0.0615 x 124

= 7.626 g

Answer:

RbF2 + Ca

Explanation:

Since Rb is located in IIA as well as Ca, Rb has ion form Rb2+

Hence CaF2 + Rb → RbF2 +Ca

Please be reminded, this answer may not be 100% correct

Taking into account the definition of density, the mass of the sample of oil with a volume of 10 cm³ and a density of 0.8 [tex]\frac{g}{cm^{3} }[/tex]  is 8 grams.

Density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance.

Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

[tex]density=\frac{mass}{volume}[/tex]

From this expression it can be deduced that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.

In this case, you know that:

Replacing in the definition of density:

[tex]0.8 \frac{g}{cm^{3} } =\frac{mass}{10 cm^{3} }[/tex]

Solving:

mass=  0.8 [tex]\frac{g}{cm^{3} }[/tex] × 10 cm³

mass= 8 g

In summary, the mass of the sample of oil with a volume of 10 cm³ and a density of 0.8 [tex]\frac{g}{cm^{3} }[/tex]  is 8 grams.

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Answer:

0.071 moles of Na₃PO₄ .

Explanation:

Given data:

Number of molecules of Na₃PO₄ = 4.3× 10²² molecules

Number of moles = ?

Solution:

1 mole contain 6.022 × 10²³ molecules

4.3× 10²² molecules × 1 mol / 6.022 × 10²³ molecules

0.71× 10⁻¹ mol

0.071 mol

The number 6.022 × 10²³ is called Avogadro number.

"It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance"

the answer is true the first one