Ask question

Ask question

All categories

Aleonysh [2.5K]

2 years ago

13

Isopropyl alcohol is a clear liquid with a density of 0.78 g/cm3. If the same small sphere (with a density of 0.88 g/cm3) is dro

pped in a beaker of isopropyl alcohol, predict whether the sphere will float or sink?

1 answer:

Burka [1]2 years ago

6 0

Ayyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

You might be interested in

Why alkali metals have low malting and bioling point

Arada [10]

Alkali metals have only one valence electron and so have low binding energy to the metallic crystal lattice. ... A lower amount of energy needed to break a bond means a lower melting/boiling point.

4 0

3 years ago

I'M GIVING 50 POINTS!!!!! NEEDS TO BE CORRECT EXPLAIN!!!!!!!!!

Virty [35]

There are 3 types of atoms in an ammonia solution, NH4OH: nitrogen, hydrogen and oxygen. So, your answer in this context is A.
However, if you were considering just ammonia on its own (NH3) there are only two types of atoms, nitrogen and hydrogen.

8 0

2 years ago

Read 2 more answers

A 150.0 mL solution of 2.888 M strontium nitrate is mixed with 200.0 mL of a 3.076 M sodium fluoride solution. Calculate the mas

Lelechka [254]

Answer:

Mass SrF2 produced = 38.63 g SrF2 produced

[Na^+]: = 1.758 M

[NO3^-]: = 1.238 M

[Sr^2+] = 0.3589 M

[F^-] = 2.36*10^-5 M

Explanation:

Step 1: Data given

Volume of 2.888M strontium nitrate = 150.0 mL = 0.150 L

Volume of 3.076 M sodium fluoride = 200.0 mL = 0.200 L

Step 2 : The balanced equation

Sr(NO3)2(aq) + 2NaF(aq) → SrF2(s) + 2NaNO3(aq) → Sr2+ + 2F- + 2

Step 3: Calculate moles strontium nitrate

Moles Sr(NO3)2 = Molarity * volume

Moles Sr(NO3)2 = 2.888 M * 0.150 L

Moles Sr(NO3)2 = 0.4332 moles

Step 4: Calculate moles NaF

Moles NaF = 3.076 M * 0.200 L

Moles NaF = 0.6152 moles

It takes 2 moles F^- to precipitate 1 mole Sr^2+, so F^- is limiting.

Step 5: Calculate limiting reactant

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

NaF is the limiting reactant. It will completely be consumed (0.6152 moles).

Sr(NO3)2 is in excess. There will react 0.6152/2 = 0.3076 moles

Moles Sr^2+ precipitated by F^- = 0.3076

There will remain 0.4332 - 0.3076 = 0.1256 moles of Sr(NO3)2

Moles Sr^2+ no precipitated (left over) = 0.1256 moles

Step 6: Calculate moles SrF2

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

For 0.6152 moles NaF we have 0.6152/2 = 0.3076 moles of SrF2

Mass SrF2 produced: 0.3076 mol * 125.6 g/mol = 38.63 g SrF2 produced

Step 7: Calculate concentration of [Na+] and [NO3-]

Since both Na^+ and NO3^- are spectator ions, and the final volume is 150 ml + 200 ml = 350 ml (0.350 L), the concentrations of Na^+ and NO3^- can be calculated as follows:

[Na^+]: (200 ml)(3.076 M) = (350 ml)(x M) and x = 1.758 M

[NO3^-]: (150 ml)(2.888 M)(2) = (350 ml)(x M) = 1.238 M

Step 8: Calculate [Sr^2+] and [F^-]

[Sr^2+] = 0.1256 moles/0.350 L = 0.3589 M

To find [F^-], one needs the Ksp for SrF2. There are several values listed in the literature. I am using a value of 2x10^-10.

SrF2(s) <==> Sr^2+(aq) + 2F^-(aq)

Ksp = [Sr^2+][F^-]²

2x10^-10 = (0.3589)(x)²

x² = 5.57*10^-10

x = [F^-] = 2.36*10^-5 M

4 0

3 years ago

When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to ef

kumpel [21]

Answer:

<em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

<em></em>

Explanation:

<em>mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.</em>

mass = 3.51 g

<em>lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec</em>

From this we can say that

rate of effusion of CO2 = 3.51/1 = 3.51 g/s

rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s

<em>From graham's equation of effusion</em>,

$\frac{Rc}{Rp}$ = $\sqrt{\frac{Mp\\}{Mc} }$

Rc = rate of effusion of CO2 = 3.51 g/s

Rp = rate of effusion of phosphorus chloride = 1.98 g/s

Mc = molar mass of CO2 = 44.01 g/mol

Mp = molar mass of the phosphorus chloride = ?

Imputing values into the equation, we have

$\frac{3.51}{1.98}$ = $\sqrt{\frac{Mp\\}{44.01} }$

1.77 = $\frac{\sqrt{Mp} }{6.64}$

11.75 = $\sqrt{Mp}$

Mp = $11.75^{2}$

Mp = <em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

5 0

3 years ago

HELPP ILL GUVE YOU LOTS OF POINTS EVERYTHING U WANT I REALLY NEED HELP

SpyIntel [72]

Answer:

jupiter is your answer

Explanation:

you weigh more there

3 0

2 years ago