What is true about an element

Suppose you begin with an unknown volume of 8.61 m h2so4 and add enough water to make 5.00*102 ml of a 1.75 m h2so4 solution. wh

olga55 [171]

1.02x10^2 ml Since molarity is defined as moles per liter, the product of the molarity and volume will remain constant as mole solvent is added. So let's set up an equality to express this m0*v0 = m1*v1 where m0, v0 = molarity and volume of original solution m1, m1 = molarity and volume of final solution. Solve for v0, then substitute the known values and calculate: m0*v0 = m1*v1 v0 = (1.75 M * 500 ml)/8.61 M v0 = (1.75 M * 500 ml)/8.61 M V0 = 101.6260163 Rounding to 3 significant figures gives 102 ml. So the original volume of the 8.61 M H2SO4 solution was 102 ml or 1.02x10^2 ml.

3 0

2 years ago

A solution of [FeSCN]^{2+} is found to have 37% transmittance at 447 nm. If the molar absorption coefficient (ε) is 4400 at λ_{m

alexandr1967 [171]

Answer:9.81×10^-5 M

Explanation: See attachments

8 0

3 years ago

Initial temperature of metal =

Evgesh-ka [11]

Answer:

the correct answers is 100 22.7 and 24.6

Explanation:

did it on edgunity

4 0

3 years ago

Read 2 more answers

Which of the following statements about benzene is true? Multiple Choice

fenix001 [56]

Answer: option a and d

Explanation:

Option A- Benzene undergoes substitution reaction

Example : benzene reacts with chlorine to form chlorobenzene, in the presence of Iron

(iii) chloride as a catalyst

C6H6 + Cl2 ---> C6H5Cl + HCl

Option D- Benzene also undergoes addition reaction

Example: benzene reacts with hydrogen , in the presence of nickel as a catalyst to form

cyclohexane

C6H6 + 3H2 ---> C6H12

Reasons why Option B isn't the answer

Although benzeme has degree of unsaturation but it's not five degree of unsaturation.

Benzene has 6 carbon atoms and 4 degrees of unsaturation (1 ring and 3 double

bonds).

If you work backwards and double the degrees of unsaturation you have 8 degrees of

unsaturation instead of 5.

Option C - Benzene isn't a saturated hydrocarbon

6 0

2 years ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

2 years ago