Volumes of 0.18M copper (II) sulfate, and water.

6.5ml 0.18M CuSO4

4.0ml H20

This is my chemistry finals, I need help immediately!

Answer:

[tex]\%m=66.7\%[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the concentration of the solution obtained, by knowing 20 mL of water are the same to 20 g and therefore the mass of the solution is 40g+20g=60g.

Next, we apply the following equation to obtain the required concentration:

[tex]\%m=\frac{40g}{60g} *100\%\\\\\%m=66.7\%[/tex]

Regards!

Various structures, or allotropes, of carbon, are precious stone, graphite, and fullerenes. In jewel, every carbon iota is attached to four other carbon iotas, shaping an unbending construction that makes precious stones hard.

Answer:

36g

Explanation:

you need to know the equation mass=moles*mr (in this case mr of carbon which is 12)

so 3*12=36g

hope this helps :)

Answer:

Group 1A: alkali metals, or lithium family.

Group 2A: alkaline earth metals, or beryllium family.

Group 7A: the manganese family.

Group 8A: the iron family.

Explanation:

Answer:

1A: Alkali Metals

2A: Alkaline Earth Metals

7A: Halogens

8A: Noble Gases

Answer:

B

Explanation:

So, a pot of boliling is hot right? of course, since it is hot thermal energy will be transferred from one place to another. I don't know if this is correct but I just wanted to give it a try.

Solution :

The equation is :

[tex]$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$[/tex]

The number of the moles of HA os 0.00285, and the volume is 25 mL.

15 mL of the 0.0950 M NaOH is added.

The total volume of a solution is V = 25 mL + 15  mL = 40 mL

The pH of the solution is 6.50

Calculating the [tex]K_a[/tex] of HA

[tex]$HA(aq) \rightleftharpoons A^-(aq)+H^+$[/tex]

[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]

Let s calculate the concentration of HA and NaOH

[tex]$[HA] = \frac{^nH_A}{V}$[/tex]

        [tex]$=\frac{0.00285 \ mol}{0.04 \ L}$[/tex]

       = 0.07125 M

[tex]$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$[/tex]

            [tex]$=\frac{0.001425 mol}{0.04L}$[/tex]

           = 0.0356 M

                                      [tex]$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$[/tex]

Initial conc. (M)            0.07125 M       0.0356 M            0 M

Change in conc. (M)   -0.0356 M       -0.0356 M        + 0.0356 M

Equilibrium conc. (M)   0.03565 M        0 M                0.0356 M

Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M

0.0356 M of NaA dissociates completely into 0.0356 M [tex]Na^+[/tex] and 0.0356 M [tex]A^-[/tex]

Now for [tex][H^+][/tex]

[tex]$[H^+] = 10^{-pH}$[/tex]

       [tex]$=10^{-6.5}$[/tex]

       [tex]$=3.16 \times 10^{-7}$[/tex]

Calculating the value of [tex]K_a[/tex],

[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]

     [tex]$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$[/tex]

     [tex]$=3.16\times 10^{-7}$[/tex]

Therefore the the value of [tex]K_a[/tex] for the unknown acid is [tex]$3.16\times 10^{-7}$[/tex].

Answer:

32.0 kJ

General Formulas and Concepts:

Thermochemistry

Specific Heat Formula: q = mcΔT

Explanation:

Step 1: Define

Identify variables

[Given] m = 1.00 g

[Given] ΔT = 1.48 °C

[Given] c = 21.6 kJ/g °C

[Solve] q

Step 2: Find Heat

Step 3: Check

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

31.968 kJ ≈ 32.0 kJ

Answer:

7

Explanation:

Answer:

See explanation and image attached

Explanation:

A Grignard reagent is an alkyl magnesium halide. If it reacts with ethyl formate, an intermediate is formed as shown.

This intermediate can undergo water hydrolysis to form a diol, ethanol and MgBrOH.

Oxidation of the diol obtained now yields the corresponding alkanal which in this case is ethanal.

The scheme of the reaction is shown in the image attached to this answer.

Answer:

50 gram calcium do you need

Explanation:

please make me brainlist answer

maybe is answer is rhyolite

Answer:

the answer is rhyolite.

Explanation:

i'm pretty sure it is my guy

Answer:

See explanation

Explanation:

Esterification is a reaction that involves the combination of an alkanoic acid and an alkanol. The product is always a sweet smelling substance.

Sulphuric acid acts as a catalyst in this reaction because it is a dehydrating agent thereby pushing the equilibrium position towards the right by the removal of water molecules.

The test tubs is heated in a water bath and not directly moved the flame because the alcohol is flammable. Also heating in a water bath helps to separate the reaction mixture from the newly formed ester.

Esters are used in industries that produces soaps and perfumes. There is a great need for the use of fragrances which are ester compounds in these industries.

The wet paper towel in the opening of the test tube cools the top of the test tube. It usually serves as a kind of condenser preventing an excess loss of vapour from the reaction mixture.

The reaction of pentanol and ethanoic acid yields pentyl ethanoate according to IUPAC nomenclature.

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

[tex]rate = k[P]^{2} [Q][/tex]

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

[tex]rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }[/tex]

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

[tex]k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4[/tex]

Second row:

2. Rate value:

[tex]rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1[/tex]

3.Third row:

[tex][Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}[/tex]

4. Fourth row:

[tex][P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}[/tex]

Answer:

49.5J/°C

Explanation:

The hot water lost some energy that is gained for cold water and the calorimeter.

The equation is:

Q(Hot water) = Q(Cold water) + Q(Calorimeter)

Where:

Q(Hot water) = S*m*ΔT = 4.184J/g°C*54.56g*(80.4°C-59.4°C) = 4794J

Q(Cold water) = S*m*ΔT = 4.184J/g°C*47.24g*(59.4°C-40°C) = 3834J

That means the heat gained by the calorimeter is

Q(Calorimeter) = 4794J - 3834J = 960J

The calorimeter constant is the heat gained per °C. The change in temperature of the calorimeter is:

59.4°C-40°C = 19.4°C

And calorimeter constant is:

960J/19.4°C =

Answer:

Consider the reaction between an alcohol and tosyl chloride, followed by a nucleophile. Write the condensed formula of the expected main organic product.

CH3CH2CH2OH---------- 2.CI 1.TsCl,pyridine__________

Explanation:

Given alcohol is propanol.

When it reacts with TsCl, the hydrogen in -OH group is replaced with tosyl group.

Pyridine is a weak base and it neutralizes the HCl (acid) formed during the reaction.

The reaction is shown below:

Answer:

the answer is b.CH3NO2 I guess I'm correct

Answer:

95.9 kg

Explanation:

First we convert 15.0 mi² to m²:

Then we convert 27.0 ft to m:

Now we calculate the total volume of the lake:

Converting 3.20x10⁸ m³ to L:

Now we calculate the total mass of mercury in the lake, using the given concentration:

Finally we convert μg to kg:

Answer:

The pressure will be 933.33 Kpa

Explanation:

Given that:

Volume V₁ = 200 cm³  (note, there is a mistake in the volume. It is supposed to be 200 cm³)

Pressure P₁ = 700 Kpa

Pressure P₂ = ??? (unknown)

Volume V₂ = 150 cm³

Temperature = constant

Using Boyle's law:

PV = constant

i.e.

P₁V₁ = P₂V₂

700 Kpa × 200 cm³ = P₂ × 150 cm³

P₂ = (700 Kpa × 200 cm³)/150 cm³

P₂ = 933.33 Kpa

Answer:

I think A should be the answer because oxygen is the chemical change of carbon.

The question is incomplete, the complete question is:

When aqueous solutions of NaCl and [tex]Pb(NO_3)_2[/tex] are mixed, a solid forms. Determine the mass of solid formed when 140.7 mL of 0.1000 M NaCl is mixed with an excess of an aqueous solution of

Answer: The mass of lead chloride produced is 1.96 g

Explanation:

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{ \text{Volume of solution (mL)}}[/tex] .....(1)

Given values:

Molarity of NaCl = 0.1000 M

Volume of the solution = 140.7 mL

Putting values in equation 1, we get:

[tex]0.1000=\frac{\text{Moles of NaCl}\times 1000}{140.7}\\\\\text{Moles of NaCl}=\frac{0.1000\times 140.7}{1000}=0.01407mol[/tex]

The chemical equation for the reaction of NaCl and lead nitrate follows:

[tex]Pb(NO_3)_2(aq)+2NaCl(aq)\rightarrow PbCl_2(s)+2NaNO_3(aq)[/tex]

By the stoichiometry of the reaction:

If 2 moles of NaCl produces 1 mole of lead chloride

So, 0.01407 moles of NaCl will produce = [tex]\frac{1}{2}\times 0.01407=0.007035mol[/tex] of lead chloride

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(2)

Molar mass of lead chloride = 278.1 g/mol

Plugging values in equation 2:

[tex]\text{Mass of lead chloride}=(0.007035mol\times 278.1g/mol)=1.96g[/tex]

Hence, the mass of lead chloride produced is 1.96 g

Answer:

Explanation:

578kj/mol

Answer:

0.189 g

Explanation:

Step 1: Write the balanced equation

NCl₃ + 3 H₂O ⇒ NH₃ + 3 HCIO

Step 2: Calculate the moles corresponding to 1.33 g of NCl₃

The molar mass of NCl₃ is 120.36 g/mol.

1.33 g × 1 mol/120.36 g = 0.0111 mol

Step 3: Calculate the moles of NH₃ produced from 0.0111 moles of NCl₃

The molar ratio of NCl₃ to NH₃ is 1:1. The moles of NH₃ produced are 1/1 × 0.0111 mol = 0.0111 mol.

Step 4: Calculate the mass corresponding to 0.0111 moles of NH₃

The molar mass of NH₃ is 17.03 g/mol.

0.0111 mol × 17.03 g/mol = 0.189 g

Answer:

Explanation:

Bromination of allylic compounds occurs when hydrogen atoms from neighboring double bonds are replaced. As a result, there are four (4) potential bromination products, as seen in the figure below.

NBS, also known as N-Bromo succinimide, is employed as a replacement for Br2 in certain instances. The benefit of NBS is that it produces a reduced level concentration for Br2, which means that bromination of the double bond isn't competitive. As soon as Br2 has been produced, the reaction continues in the same way as the remaining free-radical halogenation reactions.

Answer:i believe you are to decompose the formula (i think)

Answer:

3.59x10⁻⁴ mol

Explanation:

Assuming ideal behaviour we can solve this problem by using the PV=nRT formula, where:

We input the data given by the problem:

And solve for n:

Answer: The mass of [tex]CO_2[/tex] produced is 12.32 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of ethane = 4.21 g

Molar mass of ethane = 30 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of ethane}=\frac{4.21g}{30g/mol}=0.140mol[/tex]

Given mass of oxygen gas = 31.9 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of oxygen gas}=\frac{31.9g}{32g/mol}=0.997mol[/tex]

The chemical equation for the combustion of ethane follows:

[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]

By stoichiometry of the reaction:

If 2 moles of ethane reacts with 7 moles of oxygen gas  

So, 0.140 moles of ethane will react with = [tex]\frac{7}{2}\times 0.140=0.49mol[/tex] of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, ethane is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of ethane produces 4 moles of [tex]CO_2[/tex]

So, 0.140 moles of ethane will produce = [tex]\frac{4}{2}\times 0.140=0.28mol[/tex] of [tex]CO_2[/tex]

We know, molar mass of [tex]CO_2[/tex] = 44 g/mol

Putting values in above equation, we get:

[tex]\text{Mass of }CO_2=(0.28mol\times 44g/mol)=12.32g[/tex]

Hence, the mass of [tex]CO_2[/tex] produced is 12.32 g

Answer:

35.75 days

Explanation:

From the given information:

For first-order kinetics, the rate law can be expressed as:

[tex]\mathsf{In \dfrac{C}{C_o} = -kt}[/tex]

Given that:

the rate degradation constant = 0.12 / day

current concentration C = 0.05 mg/L

initial concentration C₀ = 3.65 mg/L

[tex]\mathsf{In( \dfrac{0.05}{3.65})= -(0.12) t}[/tex]

㏑(0.01369863014) = -(0.12) t

-4.29 = -(0.12)

t = -4.29/-0.12

t = 35.75 days