A roll of tape measures 45.5 inches. What is the length of the tape in meters?

Which option describes energy being released as heat?

Mnenie [13.5K]

Answer:

I’m pretty sure it’s Lions sleeping after a big meal

Explanation:

8 0

2 years ago

Read 2 more answers

The question is on the pic, thanks :)

Inessa05 [86]

It’s the BOA not the dog or kangaroo

8 0

1 year ago

A 0.1064 g sample of a pesticide was decomposed by the action of sodium biphenyl. The liberated Cl- was extracted with water and

Ilya [14]

Answer:

Percentage of an aldrin in the sample is 44.41%.

Explanation:

$Cl^-+AgNO_3\rightarrow AgCl+NO_{3}^-$

Molarity of the silver nitrate solution = 0.03337 M

Volume of the silver nitrate = 23.28 mL = 0.02328 L

Moles of silver nitrate = n

$0.03337 M=\frac{n}{0.02328 L}$

n = $0.03337 M\times 0.02328 L=0.0007768 mol$

According to reaction 1 mole of silver nitrate recats with 1 moles of chloride ions.

Then 0.0007768 moles of silver nitrate will react with:

$\frac{1}{1}\times 0.0007768 mol=0.0007768 mol$ chloride ions.

In one mole of aldrin there are 6 moles of chloride ions.

Then moles of aldrin containing 0.0007768 moles chloride ions are:

$\frac{0.0007768 mol}{6}=0.0001295 mol$

Moles of aldrin present in the sample = 0.0001295 mol

Mass of 0.0001295 moles of aldrin present in the sample :

0.0001295 mol × 364.92 g/mol =0.04726 g

Percentage of an aldrin in the sample:

$\frac{0.04726 g}{0.1064 g}\times 100=44.41\%$

8 0

2 years ago

A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na

Leona [35]

Answer: pH=12.69

Explanation:

${\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of HF}=0.20M\times 0.6L=0.12 moles$

$HF\rightarrow H^++F^-$

Initial 0.12 0 0

Eqm 0.12-x x x

$K_a=\frac{[H^+][F^-]}{[HF]}$

$3.5\times 10^{-4}=\frac{x^2}{0.12-x}$

(neglecting small value of x in comparison to 0.12)

$x=4.2\times 10^{-5}$

Moles of $H^+=4.2\times 10^{-5}$

$NaOH\rightarrow Na^++OH^-$

${\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles$

0.06 moles of NaOH will give 0.06 moles of $[OH^-]$

Now $4.2\times 10^{-5}$ moles of $OH^-$ will be neutralized by $4.2\times 10^{-5}$ moles of $H^+$ and $(0.06-4.2\times 10^{-5})=0.059$ moles of $OH^-$ will be left.

Molarity of $OH^-=\frac{0.059moles}{1.2L}=0.049M$

$pOH=-\log[OH^-]=-\log[0.049]=1.31$

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0

2 years ago

A 0.050 M solution of AlCl3 had an observed osmotic pressure of 3.85 atmatm at 20°C.Calculate the van't Hoff factor iii for AlCl

Alja [10]

Answer:

The actual Van't Hoff factor for AlCl3 is 3.20

Explanation:

Step 1: Data given

Molarity of AlCl3 = 0.050 M

osmotic pressure = 3.85 atm

Temperature = 20 °C

Step 2: Calculate the Van't Hoff factor

AlCl3(aq) → Al^3+(aq) + 3Cl^-(aq)

The theoretical value is 4 ( because 1 Al^3+ ion + 3 Cl- ions) BUT due to the interionic atractions the actual value will be less

Osmotic pressure depends on the molar concentration of the solute but not on its identity., and is calculated by:

π = i.M.R.T

⇒ with π = the osmotic pressure = 3.85 atm

⇒ with i = the van't Hoff factor

⇒ with M = the molar concentration of the solution = 0.050 M

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 20 °C = 293.15 Kelvin

i = π /(M*R*T )

i = (3.85) / (0.050*0.08206*293.15)

i = 3.20

The actual Van't Hoff factor is 3.20

6 0

2 years ago