A reaction was performed, and the dichloromethane solvent was dried by adding magnesium sulfate drying agent. When the reaction flask was shaken, it was observed that the magnesium sulfate clumped together at the bottom of the flask. What does this observation indicate

Answers

Answer 1

The clumping of magnesium sulfate means that the wrong kind of drying agent have been used for the sample.

What is a drying agent?

A drying agent is also referred to as a desiccant. It is a substance that is used to remove moisture from a sample. We must recall that the drying agent to be used must not react with the sample.

Since the magnesium sulfate was found to clump together at the bottom of the flask, it means that the wrong kind of drying agent have been used for the sample.

Learn more about drying agent: https://brainly.com/question/25776319


Related Questions

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is ________ M.
A) 0.0735
B) 0.0762
C) 0.0980
D) 0.0709
E) 0.00253

Answers

Answer: The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.

Explanation:

Given: Concentration of hydrogen fluoride = 0.126 M

Concentration of fluoride ions = 0.1 M

Volume of HCl = 9.0 mL

Concentration of HCl = 0.01 M

Volume of HCl = 25.0 mL

Moles of [tex]F^{-}[/tex] ions are calculated as follows.

[tex]Moles of F^{-} = molarity \times volume\\= 0.1 M \times 0.025 L\\= 0.0025 mol[/tex]

Moles of HF are as follows.

[tex]Moles of HF = Molarity \times Volume\\= 0.126 M \times 0.025 L\\= 0.00315 mol[/tex]

Moles of HCl are as follows.

[tex]Moles of HCl = Molarity \times volume\\= 0.01 M \times 0.009 L\\= 0.00009 mol[/tex]

Now, reaction equation with initial and final moles will be as follows.

                        [tex]H^{+} + F^{-} \rightarrow HF[/tex]

Initial:      0.00009  0.0025    0.00315

Equilibrium:      (0.0025 - 0.00009)    (0.00315 + 0.00009)

                                = 0.00241                    = 0.00324

Total volume = (9.00 mL + 25.0 mL) = 34.0 mL = 0.034 L

Hence, concentration of fluoride ions is calculated as follows.

[tex]Concentration = \frac{moles}{volume}\\= \frac{0.00241 mol}{0.034 L}\\= 0.0709 M[/tex]

Thus, we can conclude that concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.

Determine the type of alcohol corresponding to each given description or name. 1-pentanol 3-ethyl-3-pentanol 2-hexanol An alcohol with two other carbons attached to the carbon with the hydroxyl group_____.An alcohol with one other carbon attached to the carbon with the hydroxyl group____.An alcohol with three other carbons attached to the carbon with the hydroxyl group____.

Answers

Answer:

1). 1-pentanol - Primary

2). 3-ethyl-3-pentanol - Tertiary

3). 2-hexanol - Secondary

4). Alcohol with two other carbons attached to the carbon with the hydroxyl group - Secondary

5). Alcohol with one other carbon attached to the carbon with the hydroxyl group - Primary

6). Alcohol with three other carbons attached to the carbon with the hydroxyl group - Tertiary

Explanation:

The distinct types of alcohol have been matched with the categories above as per their descriptions provided. In chemistry, alcohols have been categorized into three different categories namely primary, secondary, and tertiary.

In the primary type, those alcohols are involved in which there is an association of hydroxyl group to a primary atom of carbon along with a minimum of two atoms of hydrogen. Example; ethanol.

In the secondary type, the alcohols have an association of carbon atoms to hydroxyl with a single atom of hydrogen and has a formula of '-CHROH.' Example: 2 - propanol.

In the tertiary alcohols, here the association is between the hydroxyl group with the carbon atom that is saturated and possessing 3 atoms of carbon associated with it. It has a formula of '-CR2OH.' Example:  3-ethyl-3-pentanol, -tert -butyl alcohol, etc.

An example of a molecular compound that obeys the octet rule in which all atoms have a zero formal charge is:

Answers

Answer:

[tex]NCl_3[/tex]

Explanation:

An octet rule is a thumb rule in the chemical sciences in which there is a natural tendency for an atom to prefer eight electrons in the valence shell of the atom. When there are less than eight electrons in the atom, they react with other atoms and form more stable compounds.

In the context, nitrogen trichloride, [tex]NCl_3[/tex], is an example of molecular compound which obeys the octet rule having a zero formal charges on each atom of the compound.

What is the difference between acids and strong acids?
a. Strong acids produce all of their H30+ ions
b. Strong acids release all of their H30+ ions
c. Strong acids produce all of their OH- ions
d.Strong acids release all of their OH ions

Answers

Answer:

A and b are the both the answers

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!

3) Convert 0.250 moles of aluminum sulfate to grams.

4) Convert 2.70 grams of ammonia to moles.

Answers

Answer:

0.000731 grams aluminium sulfate

46.0 mols ammonia

Explanation:

ALS = shorthand for aluminium sulfate which has a molar mass of 342.15 g/mol

[tex]ALS: \frac{0.250mols}{1} *\frac{1g}{342.15mols} = \frac{0.250g}{342.15}=0.0007307 g[/tex]

NH3 has a molar mass of 17.031 g/mol

[tex]NH3: \frac{2.70g}{1} *\frac{17.031mols}{1g} = \frac{0.250g}{342.15}=45.9837 mols[/tex]

Aluminium sulphate (AlS) whose molar mass is= [tex]\sf{ 342.15\dfrac{g}{mol} }[/tex]

we have to find the 0.250 moles of aluminum sulphate.

[tex]\implies AlS=\dfrac{1g}{342.15~mole}×0.250~mole \\\\\implies AlS=\dfrac{0.250}{342.15}\\\\\implies \dfrac{\frac{250}{1000}}{\frac{34215}{100}}\\\\\implies \dfrac{250}{1000}×\dfrac{100}{34215}\\\\=0.00073067\approx{0.0007307~g}[/tex]

[tex]\\\\\\[/tex]

Ammonia(NH3) whose molar mass is =[tex]\sf{17.031\dfrac{mol}{g} }[/tex]

We have to find 2.70 grams of ammonia

[tex]\implies NH_{3}=\dfrac{17.031~mol}{1g}×2.70g\\\\ 17.031×2.70\\\\\dfrac{17031}{1000}×\dfrac{270}{100}\\\\ \dfrac{4598370}{100000}\\\\=45.9837\approx{46~mole}[/tex]

A gaseous mixture containing 7.00 moles of nitrogen, 2.50 moles of oxygen, and 0.500 mole of helium exerts a total pressure of 0.900 atm. What is the partial pressure of the nitrogen?

Answers

Answer:

Partial Pressure = 0.630atm

Explanation:

The total pressure of a mixture of gases is equal to the partial pressure of those gases. The partial pressure of a gas is defined as:

Partial pressure = Mole Fraction * Total pressure

The mole fraction of a gas is the ratio between the moles of the gas and the total moles.

To solve this question we need to find the mole fraction of nitrogen to find its partial pressure:

Mole Fraction nitrogen:

7.00 moles Nitrogen / (7.00moles N2 + 2.50moles O2 + 0.500moles He) = 0.700 = Mole fraction.

Partial Pressure = 0.700* 0.900atm

Partial Pressure = 0.630atm

For a given fluorophore, select the choice that correctly lists the processes of fluorescence, absorption, and phosphorescence in order from shortest to longest wavelength.

a. absorption, fluorescence, phosphorescence
b. Fluorescence = phosphorescence, absorption
c. fluorescence, phosphorescence, absorption
d. phosphorescence, fluorescence, absorption
e. absorption, phosphorescence, fluorescence
f. absorption, fluorescence = phosphorescence

Answers

Answer:

absorption, fluorescence = phosphorescence

Explanation:

Given a particular fluorophore, the wavelength of absorption of the fluorophore is always shorter. Both fluorescence and phosphorescence are kinds of photoluminescence.

Recall that both fluorescence and phosphorescence occur at a longer wavelength. The difference between the two is only in the time taken during the process. While fluorescence takes a shorter time to occur, phosphorescence takes a longer time to occur.

The major difference between fluorescence and phosphorescence is that change of spin occurs during phosphorescence but not fluorescence.

Fructose is an example of a ketohexose. The -hexose part of the name indicates that fructose is a Choose... that contains Choose... carbons. The keto- part of the name indicates that fructose contains Choose... functional group. Fructose can combine with glucose to form sucrose. Therefore, sucrose is a Choose... .

Answers

Answer:

carbohydrate, 6, a carbonyl, disaccharide

Explanation:

Fructose is an example of a ketohexose. The -hexose part of the name indicates that fructose is a carbohydrate that contains 6 carbons. There are more isomers that are ketohexoses.

The keto- part of the name indicates that fructose contains a carbonyl functional group. In ketones, the carbonyl is in an inner carbon.

Fructose can combine with glucose to form sucrose. Therefore, sucrose is a disaccharide. Disaccharides are formed by the bonding of 2 monosaccharides.

Help naming this plzzzzzzzzzzzzz

Answers

Answer:

A. 3-chloro-1-methylcyclobutane.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to infer that the name of this compound is A. 3-chloro-1-methylcyclobutane because of the fact that the parent chain is a cyclobutane which starts by the methyl radical as it has the priority over the chlorine radical which is actually named first at the third carbon (clockwise).

Therefore the name is given in A, accordingly to the IUPAC rules of nomenclature.

Regards!

Part A
If the theoretical yield of a reaction is 23.5 g and the actual yield is 14.8 g, what is the percent yield?

Answers

Answer:

[tex]\boxed {\boxed {\sf 63.0 \%}}[/tex]

Explanation:

The percent yield is the ratio of the actual yield to the theoretical yield.

[tex]percent \ yield = \frac{actual \ yield}{theoretical \ yield} * 100[/tex]

The actual yield is the amount obtained from performing a chemical reaction. For this problem, it is 14.8 grams. The theoretical yield is the potential amount from performing a chemical reaction at maximum performance. For this problem, it is 23.5 grams.

We can substitute the known values into the formula.

[tex]percent \ yield= \frac{ 14.8 \ g}{23.5 \ g}*100[/tex]

Divide.

[tex]percent \ yield = 0.629787234043*100[/tex]

Multiply.

[tex]percent \ yield = 62.9787234043[/tex]

The original measurements for the theoretical and actual yields have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place.

The 7 to the right, in the hundredths place, tells us to round the 9 up to a 0. Since we rounded up to 0, we have to move to the next place to the left and round the 2 up to a 3.

[tex]percent \ yield \approx 63.0[/tex]

The percent yield is approximately 63.0 percent.

Identify the most and the least acidic compound in each of the following sets. Leave the remaining answer in each set blank. a) difluoroacetic acid: _______ fluoroacetic acid: _______ trifluoroacetic acid: _______ b) 2-chlorobutanoic acid: _______ 4-chlorobutanoic acid: _______ 3-chlorobutanoic acid: _______ c) cyclohexanol: _______ phenol: _______ benzoic acid:

Answers

Explanation:

The given set of options are:

a) difluoroacetic acid: _______ fluoroacetic acid: _______ trifluoroacetic acid: _______

b) 2-chlorobutanoic acid: _______ 4-chlorobutanoic acid: _______ 3-chlorobutanoic acid: _______

c)cyclohexanol: _______ phenol: _______ benzoic acid:

A strong acid is one whose conjugate base is stabilized either by resonance or -I effect.

a) If -I groups are present on the carboxylic acid group then they stabilize the carboxylate anion (the conjugate base of the carboxylic acid) and give more strength to the carboxylic acid group. More the number of -I groups then more will be the strength of the carboxylic acid.

Among the given options,

Trifluoroacetic acid has three fluorine atoms in its structure thus it is a strong acid.

Fluoroacetic acid has only one fluorine atom in its structure. Hence it is the weak acid among the given options.

b) The -I groups should be nearer to the carboxylic acid group then it attains more stability.

If the distance of the -I group increases from the -COOH group then, the strength of the carboxylic acid group decreases.

So, the strongest acid is 2-chlorobutanoic acid.

The weak acid among them is 4-chlorobutanoic acid.

c) Among the given options benzoic acid is the strongest acid because due to resonance benzoate anion is stabilized more.

Here resonance exists in both phenyl group and [tex]-COO^-[/tex] group.

The weak acid is cyclohexanol.

Because cyclohexanolate anion is not stabilized by resonance.

Is sucrose classified as aldose or ketose?

Answers

Answer:

Because sucrose is a complex disaccharide, it is not classified as either an aldose or a ketone. Instead, it is a compound that contains both. glucose is aldose sugar and fructose is a ketose sugar.

Calcium carbonate reacts with stomach acid (HCI) according to the following equation:
Caco, (s) + 2 HCl(aq) - CaCl, (aq) + CO2(g) + H20 (1)
Tums, an antacid, contains CaCo3. If Tums is added to 25.7 mL of a 0.738 M HCl solution, how many grams of Co, are produced?

Answers

Answer:

0.418 g of CO₂ will be produced

Explanation:

Equation of the reaction is given below:

CaCO₃ (s) + 2HCl (aq) ----> CaCl₂(aq) + CO₂ (g) + H₂O (l)

From the equation of reaction , 1 mole of CaCO₃ reacts with 2 molemolesmof HCl to produce 1 mole of CO₂

Number of moles of HCl in 25.7 ml of a 0.738 M solution of HCl is obtained using the formula below:

Number of moles = molarity × volume (Litres)

Number of moles of HCl = 0.738 M × 25.7 mL × 1 L/1000 mL = 0.0190 moles

Since the antacid, Tums will be excess, the limiting reactant is HCl

2 moles of HCl produces 1 mole of CO₂

0.0190 moles of HCl will produce 0.0190 moles ÷ 2 = 0.0095 molesof CO₂

Mass of 0.0095 moles of CO₂ = Numbe of moles × molar mass of CO₂

Molar mass of CO₂ = 44.0 g/mol

Mass of CO₂ produced = 0.00950 moles × 44.0 g/mol = 0.418 g of CO₂

Therefore, 0.418 g of CO₂ will be produced if Tums is added to 25.7 mL of a 0.738 M HCl solution

办理教留服学位学历认证Q/微29304199英属哥伦比亚UBC毕业证文凭学位证书offer操办英属哥伦比亚留信认证成绩单

Answers

Answer:

please translate in english

time is direct propotinally to rate of chemical reaction .explain if time is negretted and temperature remain costant?​

Answers

Answer:

the constant temperature will be 12 .F bec in places it is cold

Hi,Valency of chlorine is 1. Why?Thank you​

Answers

Answer:

The chlorine element belongs to group 17 because it has 7 valence electron . Its valency is 1 . It can gain one electron from any other atom to become stable. This means that it can never form a double or triple bond.

Question 2 10
10 Points
Which of the following chemical equations depicts a balanced chemical equation?
O A. AgNO, Kcro > KNO, Agro,
OB. AgNO3 + Kycro, » 2K NO; + Agro,
C.3AgNO3 + 2K,Cro--> 3KNO3 + 249900,
D. 2AgNO, K Cro-> 2KNO; 1Cro,
Resol Selection

Answers

Answer:

2AgNO, K Cro-> 2KNO; 1Cro,

which of these is not a gas? A. hydrogen B. gravity C. Oxygen D. heluim

Answers

Answer:

B. Gravity

Explanation:

Gravity is a force, not a gas :)

Answer:

B: Gravity

Explanation:

gravity pulls on the gad

cho một thực phẩm có độ ẩm tương đối là 81 hỏi hoạt độ nước trong thực phẩm đó là bao nhiêu

Answers

Explanation:

For a food with a relative humidity of 81, what is the water activity in the food?

Water activity in a food = relative humidity / 100

Substitute the given values in this formula to get water activity:

Water activity in a food = 81 / 100

Hence, water activity =0.81

The compound chromium(II) chloride is a strong electrolyte. Write the transformation that occurs when solid chromium(II) chloride dissolves in water. Be sure to specify states such as (aq) or (s).

Answers

Answer:

CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)

Explanation:

Chromium (II) chloride is a strong electrolyte, that is, when dissolved in water, it completely dissociates into the ions. The cation is chromium (II) and the anion is chloride. The balanced equation for the solution of chromium (II) chloride is:

CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)

write down any two test for the CH2=CH2[ethene]​

Answers

Answer:

a) Linear polymerization

b) cyclic polymerization

When you burn a birthday candle, you may wonder whether the fire you see is matter. The
flame consists of hot, glowing gases, such as oxygen, carbon dioxide, water vapor, and
parts of the candle that have been vaporized. The heat and light given off are forms of
energy. The smoke contains ash and unburned particles.
a. From the above description of a candle flame, list at least three things that are matter and
three things that are not mattel. (6 points)
Matter
Not Matter

Answers

Answer:

matter

1. Candle. not matter

1. light

2. Unburned Particles

2. heat

3. Ash

3. energy

Explanation:

1.The candle is experiencing a solid phase into a gas phase because the heat given off causes smoke.

2. Chemical Potential Energy to Heat EnergyThe candle has Chemical Potential Energy then when it gets lit by the flame heat energy is released.

If you reacted 88.9 g of ammonia with excess oxygen, what mass of water would you expect to make? You will need to balance the equation first.

NH3(g) + O2(g) -> NO(g) + H2O(g)

Answers

Explanation:

here's the answer to your question

The combustion of hydrogen-oxygen mixtures is used to produce very high temperatures (ca. 2500 °C) needed for certain types of welding operations. Consider the reaction to be

H₂(g) + ½O₂(g) → H₂O(g) ∆H° = -241.8 kJ/mol

What is the quantity of heat evolved, in kilojoules, when a 180 g mixture containing equal parts of H₂ and O₂ by mass is burned?
in kj

Answers

Answer:

1360kJ are evolved

Explanation:

When 1mole of H2 reacts with 1/2 moles O2 producing 1 mole of water and 241.8kJ.

To solve this question we need to find the limiting reactant knowing were added 90g of H2 and 90g of O2 as follows:

Moles H2 -Molar mass: 2g/mol-

90g H2 * (1mol / 2g) = 45 moles

Moles O2 -Molar mass: 32g/mol-

90g * (1mol / 32g) = 2.81moles

For a complete reaction of 2.81 moles of O2 are needed:

2.81 moles O2 * (1mol H2 / 1/2 mol O2) = 5.62 moles H2

As there are 45 moles, H2 is the excess reactant and O2 the limiting reactant.

As 1/2 moles O2 produce 241.8kJ, 2.81 moles will produce:

2.81 moles O2 * (241.8kJ / 1/2moles O2) =

1360kJ are evolved

The quantity of heat evolved when 180 g mixture containing equal parts of H₂ and O₂ burned is

The equation for the combustion of hydrogen is given as:

[tex]\mathbf{H_2 + \dfrac{1}{2}O_2 \to H_2O \ \ \ \ \Delta H_r^0 = -241.8\ kJ/mol}[/tex]

Recall that:

number of moles  = mass/molar mass:

Since the mass of 180 g is equally shared by H₂ and O₂, then:

mass of H₂ = 90 gmass of O₂ = 90 g

The number of moles of the reactant can be determined as follows:

For H₂:

[tex]\mathbf{no \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]

[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{2.016 \ g/mol}}[/tex]

no of moles = 44.6 mol

For O₂:

[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{32 \ g/mol}}[/tex]

no of moles = 2.8 mol

Here, since O₂ has a lesser amount of mole, then O₂ is regarded as the limiting reagent here:

If 1/2 moles of O₂ produces -241.8 kJ/mol of water;

Then, the quantity of heat that will evolve when 180 g mixture containing equal parts of H₂ and O₂ burned is:

[tex]\mathbf{= \Big (\dfrac{2.81 \ mol}{\dfrac{1}{2 } \ mol }\Big) \times (-241.8 \ kJ) }[/tex]

[tex]\mathbf{= \Big (5.62\Big) \times (-241.8 \ kJ) }[/tex]

= - 1358.91 kJ

≅ - 1360 kJ

Therefore, we can conclude that the quantity of heat evolved is - 1360 kJ

Learn more about the quantity of heat here:

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The number of moles of H in 5 moles of glucose (C6H1206) is

Answers

Answer:

60 mols hydrogen

Explanation:

1 mol of glucose contains 12 mols of Hydrogen

5 mols of glucose contains 12 * 5 = 60 moles of hydrogen.

what is Lewis acid and Lewis base? give examples​

Answers

Explanation:

example is copper iron...........

300g de acido comercial se disuelve en agua destilada contenidos en un cono, cuyo radio es de 0.005Km y 300cm de altura, si la densidades de 1.2g/m3 ¿Cuál es la concentración expresada en %m/m?

Answers

Answer:

97.0%m/m es la concentración de la solución

Explanation:

El porcentaje masa/masa (%m/m) se define como 100 veces el radio entre la masa de soluto (300g de HCl) y la masa de la solución. Para hallar la masa de la solución debemos hallar la masa del agua (Solvente) haciendo uso de la ecuación del volumen de un cono. Con el volumen del cono podemos hallar la masa del agua haciendo uso de la densidad, así:

Volumen:

Volumen Cono = π*r²*h / 3

Donde r es el radio = 0.300m

h la altura = 5m

Volumen = π*(5m)²*0.300m / 3

7.85m³

Masa Agua:

7.85m³ * (1.2g / m³) = 9.42g Agua

Masa solución:

300g HCl + 9.42g Agua = 309.42g Solución

%m/m:

300g HCl  / 309.42g * 100 =

97.0%m/m es la concentración de la solución

The mass of a crucible and lid is 23.422 g. After adding a sample of hydrate compound the crucible, cover, and contents weigh 24.746 g. After heating with a Bunsen burner to remove the water of hydration, the mass of the crucible, cover, and sample was 24.213 g. How many moles of water did the hydrate compound contain

Answers

Answer:

0.030 mole

Explanation:

Mass of crucible + lid = 23.422 g

Mass of crucible + lid + compound = 24.746 g

Mass of crucible + lid + compound - water = 24.213

Mass of water = Mass of crucible + lid + compound + heat

       = 24.746 - 24.213

                = 0.533 g

Mole of water in the hydrated compound = mass of water in the compound/molar mass of water

    = 0.533/18

         = 0.0296 mole = 0.030 mole

complete the following steps.
Remember to follow lower numbered rules first.
K2S(aq) + CO(NO3)2(aq) COS (?) + KNO3 (?)
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 equatibn
for the reaction. If no reaction occurs, write “no
reaction.” (1 pt)

Answers

Answer:

See explanation

Explanation:

a) The balanced reaction equation is;

K2S + CO(NO3)2 ------> COS + 2 KNO3

b) When we include the states of matter, we have;

K2S(aq) + CO(NO3)2(aq) ----> COS(s) + 2 KNO3(aq)

c) The complete ionic equation is;

2K^+(aq) + S^2-(aq) + Co^2+(aq) + 2NO3^-(aq) ----> CoS(s) + 2K^+(aq) + 2NO3^-(aq)

Net ionic equation;

Co^2+(aq) + S^2-(aq) ----> CoS(s)

Curium – 245 is an alpha emitter. Write the equation for the nuclear reaction and identify the product nucleus.

Answers

Answer:

Please find the complete solution in the attached file.

Explanation:

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