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3 years ago

Question 4 (2 points)

Chemistry

1 answer:

dmitriy555 [2]3 years ago

7 0

Answer:

10.4 hours

Explanation:

Now from;

0.693/t1/2 = 2.303/t log (No/N)

t1/2= half life of the radioactive isotope

t = age of the radioactive isotope

No= Original amount of radioactive isotope

N= amount of radioactive isotope left after time t

Substituting values;

N = 0.0625No

0.693/2.6 = 2.303/t log (No/0.0625No)

0.693/2.6 = 2.303/t log (1/0.0625)

0.267 = 2.303/t * log 16

0.267 = 2.773/t

t = 2.773/0.267

t = 10.4 hours

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Step2247 [10]

Answer:

a)0.67 molal

b) 5.67 molal

c) 2.82 molal

d) 0.0268 molal

Explanation:

Step 1: Data given

Molality = moles solute / mass solvent

(a) 0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg of water—a saturated solution at 0°C

Calculate moles of Na2CO3 = 710 grams / 105.99 g/mol = 6.70 moles

Molality = 6.70 moles / 10 kg = 0.67 molal

(b) 125 g of NH4NO3 in 275 g of water—a mixture used to make an instant ice pack

Calculate moles of NH4NO3 = 125 grams / 80.04 g/mol = 1.56 moles

molality = 1.56 moles / 0.275 kg = 5.67 molal

(c) 25 g of Cl2 in 125 g of dichloromethane, CH2Cl2

Calculate moles Cl2 = 25 grams / 70.9 g/mol= 0.353 moles

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Calculate moles histamine = 0.372 grams / 111.15 g/moles = 0.00335 moles

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4 years ago

How are compounds Similar to elements and how are they different

Yuki888 [10]

Answer:

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3 0

3 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:

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

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3 years ago

How many oxygen molecules are required for glycolysis

morpeh [17]

Answer:

Explanation:

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6 0

3 years ago

Read 2 more answers

What is the percent composition of aluminum in the ore bauxite (AI2O3)?

Schach [20]

Answer:

Molar mass of Al2O3 = 101.961276 g/mol

This compound is also known as Aluminium Oxide.

Convert grams Al2O3 to moles or moles Al2O3 to grams

Molecular weight calculation:

26.981538*2 + 15.9994*3

Percent composition by element

Element Symbol Atomic Mass # of Atoms Mass Percent

Aluminium Al 26.981538 2 52.925%

Oxygen O 15.9994 3 47.075%

Explanation:

Percent composition by element

Element Symbol Mass Percent

Aluminium Al 52.925%

Oxygen O 47.075%

6 0

3 years ago