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A NaOH (aq) stock solution was created by dissolving 3.88 g NaOH in water to create a 100.00 mL solution. What is the concentrat

Ierofanga [76]

Answer:

I have don 3 so far

Explanation:

Brainlyis the best

8 0

2 years ago

In a desert, soil containing a mixture of sand and small rocks is exposed to wind erosion. Over time, how would the land surface

amid [387]

Sand dunes would be created due to the mixture falling on each other
x

3 0

3 years ago

Read 2 more answers

What is the osmotic pressure, in torr, of a 3.00% solution of NaCl in water when the temperature of the solution is 45 ºC? Enter

Anit [1.1K]

213034 torr is the osmotic pressure.

Explanation:

osmotic pressure is calculated by the formula:

osmotic pressure= iCrT

where i= no. of solute

c= concentration in mol/litre

R= Universal Gas constant

T = temp

It is given that solution is 3% which is 3gms in 100 ml.

let us calculate the concentration in moles/litre

3gm/100ml*1000ml/1L*1mol NaCl/55.84g NaCl

= 5.372 gm/litre

Putting the values in the formula, Temp in Kelvin 318.5K

osmotic pressure= 2*5.372*0.083 * 318.5 Gas constant 0.083

= 284.023 bar or 213018 torr. c= 5.372 moles/L

i=2 for NaCl

4 0

3 years ago

A solution contains one or more of the following ions: Ag , Ca2 , and Co2 . Lithium bromide is added to the solution and no prec

koban [17]

Answer:

$Ca^{2+} and Co^{2+}$

Explanation:

Given:

A solution contains one or more of the following ions such as Ag, $Ca_2$ and $Co_2$

Here the Lithium bromide is added to the solution and no precipitate forms

Solution:

Since with LiBr no precipitation takes place therefore Ag+ is absent

Here on adding $Li_2SO_4$ to it precipitation takes place.

Precipitate is as follows,

$Ca^{2+}(aq)+SO_4^2^-(aq)----->CASO_4(s)$

Thus,

$Ca^2^+$ is present

When $Li_3PO_4$ is added again precipitation takes place.

Therefore the reaction is as follows,

$Co^2^+(aq)+PO_4^3^-(aq)------>Co_3(PO_4)_2(s)$

Therefore,

$Ca^{2+} and Co^{2+}$ are present in the solution

3 0

3 years ago

Dissolve 30 g of sodium sulphate into 300 mL of water

Aneli [31]

Answer:

number of moles = 0.21120811

Explanation:

To find the number of moles, given the mass of the solute, we use the formula:

$\mathrm{n = \dfrac{ m }{ M } }$

$\mathrm{n = number\:of\:moles\:(mol)}$

$\mathrm{m = mass\:of\:solute\:(g)}$

$\mathrm{M = molar\:mass\:of\:solute\:( \dfrac{ g }{ mol } )}$

Label the variables with the numbers in the problem:

$\mathrm{n =\:?}$

$\mathrm{m =30\:g }$

$\mathrm{M =\:?\:Calculate\:the\:molar\:mass }$

The first thing we have to do is find the molar mass of sodium sulfate, in order for us to use the formula for finding the number of moles:

Formula for finding the molar mass of sodium sulfate:

$M({ \left Na \right }_{ 2 } { \left So \right }_{ 4 }) = m \left( Na \right) +m \left( S \right) +m \left( O \right)$

For the variables and what they mean are below for finding the molar mass of sodium sulfate:

$\mathrm{M =molar\:mass }$

$\mathrm{m =moles=2\:moles\:for\:Na\:,1\:mole\:for\:S,\:and\:4\:moles\:for\:O}$

$\mathrm{Na =sodium=22.99\:g }$

$\mathrm{S =sulfur=32.06\:g }$

$\mathrm{O =oxygen=16.00\:g }$

Plug the numbers into the formula, to find the molar mass of sodium sulfate:

$M({ \left Na \right }_{ 2 } { \left So \right }_{ 4 }) = m \left( Na \right) +m \left( S \right) +m \left( O \right)$

$\mathrm{Substitute\:the\:values\:into\:the\:formula}$

$M = 2 \left( 22.99 \right) +1 \left( 32.06 \right) +4 \left( 16.00 \right)$

$\mathrm{Multiply\:2\:by\:22.99\:to\:get\:45.98\:and\:1\:by\:32.06\:to\:get\:32.06}$

$\mathrm{M = 45.98+32.06+4\:(16)}$

$\mathrm{Multiply\:4\:by\:16\:to\:get\:64}$

$\mathrm{M = 45.98+32.06+64}$

$\mathrm{Add\:45.98\:and\:32.06\:to\:get\:78.04}$

$\mathrm{M = 78.04+64}$

$\mathrm{Add\:78.04\:and\:64\:to\:get\:142.04}$

$\mathrm{M = 142.04}$

Now that we have found the molar mass, we can calculate the number of moles in the solution of sodium sulfate with the formula:

$\mathrm{n = \dfrac{ m }{ M } }$

$\mathrm{n =\:?}$

$\mathrm{m =30\:g }$

$\mathrm{M = 142.04\:g/mol}$

$\mathrm{Substitute\:the\:values\:into\:the\:formula}$

$\mathrm{n = \dfrac{ 30 }{ 142.04 }}$

$\mathrm{Divide\:142.04\:by\:30\:to\:get\:0.21120811}$

$\mathrm{n = 0.21120811}$

0.21120811 rounded gives you 0.2112

or if you did the problem without decimals

30 grams of sodium sulfate divided by its molecular weight – which we found to be 142 – gives us a value of 0.2113 moles.

3 0

2 years ago