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frutty [35]
3 years ago
9

What are two examples of a soulution?

Chemistry
1 answer:
Anni [7]3 years ago
3 0
Which type of solution are you talking about?
You might be interested in
What is an example of conduction
Alexandra [31]

Explanation:When the stove turns on the skillet will become very hot because of the conduction of heat transferring from the stove to the skillet.

7 0
3 years ago
The concentration of a saturated BaCl2 solution is 1.75 M (mol/liter) and the concentration of a saturated Na2SO4 solution is 2.
Kitty [74]

Answer:

a) The theoretical yield is 408.45g of BaSO_{4}

b) Percent yield = \frac{realyield}{408.45g}*100

Explanation:

1. First determine the numer of moles of BaCl_{2} and Na_{2}SO_{4}.

Molarity is expressed as:

M=\frac{molessolute}{Lsolution}

- For the BaCl_{2}

M=\frac{1.75molesBaCl_{2}}{1Lsolution}

Therefore there are 1.75 moles of BaCl_{2}

- For the Na_{2}SO_{4}

M=\frac{2.0moles[tex]Na_{2}SO_{4}}{1Lsolution}[/tex]

Therefore there are 2.0 moles of Na_{2}SO_{4}

2. Write the balanced chemical equation for the synthesis of the barium white pigment, BaSO_{4}:

BaCl_{2}+Na_{2}SO_{4}=BaSO_{4}+2NaCl

3. Determine the limiting reagent.

To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:

- For the BaCl_{2}:

\frac{1.75}{1}=1.75

- For the Na_{2}SO_{4}:

\frac{2.0}{1}=2.0

As the BaCl_{2} is the smalles quantity, this is the limiting reagent.

4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

1.75molesBaCl_{2}*\frac{1molBaSO_{4}}{1molBaCl_{2}}*\frac{233.4gBaSO_{4}}{1molBaSO_{4}}=408.45gBaSO_{4}

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:

Percent yield = \frac{realyield}{theoreticalyield}*100

Percent yield = \frac{realyield}{408.45g}*100

The real yield is the quantity of barium white pigment you obtained in the laboratory.

7 0
3 years ago
Help thank you so so much!
Otrada [13]

Answer:

D is the answer to this problem

8 0
3 years ago
How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and
Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)

The expression used will be:  

Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]

where,

Q = heat released for the reaction = ?

m = mass of benzene = 94.4 g

c_{p,s} = specific heat of solid benzene = 1.51J/g^oC=1.51J/g.K

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC=1.73J/g.K

\Delta H_{fusion} = enthalpy change for fusion = -9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g

Now put all the given values in the above expression, we get:

Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]

Q=-29427.312J=-29.4kJ

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

3 0
3 years ago
Sodium reacts with chlorine gas according to the following reaction: 2Na(s)+Cl2(g)→2NaCl(s) What volume of Cl2 gas, measured at
asambeis [7]

Answer:6.719Litres of Cl2 gas.

Explanation:According to eqn of rxn

2Na +Cl2=2NaCl

P=689torr=689/760=0.91atm

T=39°C+273=312K

according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl

But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW

MW of NaCl=23+35.5=58.5g/mol

n=28g(mass given of NaCl)/58.5

n=0.479moles of NaCl

Going back to the reaction,

if 1moles of Cl2 produces 2moles of NaCl

x moles of Cl2 will give 0.479moles of NaCl.

x=0.479*1/2

x=0.239moles of Cl2.

To find the volume, we use ideal ggas eqn,PV=nRT

V=nRT/P

V=0.239*0.082*312/0.91

V=6.719Litres

6 0
3 years ago
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