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Inga [223]
3 years ago
15

3. If you mix 30mL of a 6% solution and 90mL of a 30% solution, what will the

Chemistry
1 answer:
natita [175]3 years ago
4 0

Answer: it would be 30% solution and 6% mixture (correct me if I'm wrong)

You might be interested in
An element X combines with oxygen to form a compound of formula XO2. If 24.0 grams of element X combines with exactly 16.0 grams
tensa zangetsu [6.8K]

Answer:

atomic mass of X is 48.0 amu

Explanation:

Let y be the atomic mass of X

Molar mass of O_2 is = 2×16 = 32 g / mol

X + O2 -----> XO_2

According to the equation ,

y g of X reacts with 32 g of O_2

24 g of X reacts with Z g of O_2

Z = ( 32×24) / y

But given that 24.0 g of X exactly reacts with 16.0 g of O_2

So Z = 16.0

⇒ (32×24) / y = 16.0

⇒ y = (32×24) / 16

y= 48.0

So atomic mass of X is 48.0 amu

4 0
3 years ago
What is the molar out of a solution that contains 33.5g of CaCl2 in 600.0mL of water
omeli [17]

Answer:

Here's what I got.

Explanation:

Interestingly enough, I'm not getting

0.0341% w/v

either. Here's why.

Start by calculating the percent composition of chlorine,

Cl

, in calcium chloride, This will help you calculate the mass of chloride anions,

Cl

−

, present in your sample.

To do that, use the molar mass of calcium chloride, the molar mass of elemental chlorine, and the fact that

1

mole of calcium chloride contains

2

moles of chlorine atoms.

2

×

35.453

g mol

−

1

110.98

g mol

−

1

⋅

100

%

=

63.89% Cl

This means that for every

100 g

of calcium chloride, you get

63.89 g

of chlorine.

As you know, the mass of an ion is approximately equal to the mass of the neutral atom, so you can say that for every

100 g

of calcium chloride, you get

63.89 g

of chloride anions,

Cl

−

.

This implies that your sample contains

0.543

g CaCl

2

⋅

63.89 g Cl

−

100

g CaCl

2

=

0.3469 g Cl

−

Now, in order to find the mass by volume percent concentration of chloride anions in the resulting solution, you must determine the mass of chloride anions present in

100 mL

of this solution.

Since you know that

500 mL

of solution contain

0.3469 g

of chloride anions, you can say that

100 mL

of solution will contain

100

mL solution

⋅

0.3469 g Cl

−

500

mL solution

=

0.06938 g Cl

−

Therefore, you can say that the mass by volume percent concentration of chloride anions will be

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the solution.

.

ALTERNATIVE APPROACH

Alternatively, you can start by calculating the number of moles of calcium chloride present in your sample

0.543

g

⋅

1 mole CaCl

2

110.98

g

=

0.004893 moles CaCl

2

To find the molarity of this solution, calculate the number of moles of calcium chloride present in

1 L

=

10

3

mL

of solution by using the fact that you have

0.004893

moles present in

500 mL

of solution.

10

3

mL solution

⋅

0.004893 moles CaCl

2

500

mL solution

=

0.009786 moles CaCl

2

You can thus say your solution has

[

CaCl

2

]

=

0.009786 mol L

−

1

Since every mole of calcium chloride delivers

2

moles of chloride anions to the solution, you can say that you have

[

Cl

−

]

=

2

⋅

0.009786 mol L

−

1

[

Cl

−

]

=

0.01957 mol L

−

This implies that

100 mL

of this solution will contain

100

mL solution

⋅

0.01957 moles Cl

−

10

3

mL solution

=

0.001957 moles Cl

−

Finally, to convert this to grams, use the molar mass of elemental chlorine

0.001957

moles Cl

−

⋅

35.453 g

1

mole Cl

−

=

0.06938 g Cl

−

Once again, you have

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

In reference to the explanation you provided, you have

0.341 g L

−

1

=

0.0341 g/100 mL

=

0.0341% m/v

because you have

1 L

=

10

3

mL

.

However, this solution does not contain

0.341 g

of chloride anions in

1 L

. Using

[

Cl

−

]

=

0.01957 mol L

−

1

you have

n

=

c

⋅

V

so

n

=

0.01957 mol

⋅

10

−

3

mL

−

1

⋅

500

mL

n

=

0.009785 moles

This is how many moles of chloride anions you have in

500 mL

of solution. Consequently,

100 mL

of solution will contain

100

mL solution

⋅

0.009785 moles Cl

−

500

mL solution

=

0.001957 moles Cl

−

So once again, you have

0.06938 g

of chloride anions in

100 mL

of solution, the equivalent of

0.069% m/v

.

Explanation:

i think this is it

8 0
3 years ago
In which layer(s) would a solid cube with 6 cm sides and mass of 270 g float? Explain.
DIA [1.3K]
The buoyancy of an object is dictated by its density. So let us calculate for density, where:density = mass / volume
Calculate the volume first of a solid cube:volume = (6 cm)^3 = 216 cm^3 = 216 mL
Therefore density is:density = 270 g / 216 mLdensity = 1.25 g / mL
Therefore this object will float in the layer in which the density is more than 1.25 g / mL.
6 0
3 years ago
Does the volume of particles affect the behavior of gas
lilavasa [31]

Answer:

Yes, it does, although only physically and not chemically.

Explanation:

If a volume of gas is way spread out, it won't collide with the other gas particles as often, reducing pressure and temperature because they lose kinetic energy to their surroundings when they don't collide.

If it is compressed, it increases temperature and pressure because the gas particles collide with each other and the walls of the container way more often than if they had more space.

Hope this answers your question.

P.S.

Fun fact, gas particles are actually moving at 300-400 meters per second at room temperature, they only slow down to walking speed at very low temperatures, like 10 Kelvin

7 0
3 years ago
OS-182 has a half-life of 3.6 days. If a sample started at 175 grams, how much would be led after 18 days?
hodyreva [135]
I think c is correct answer
.............
3 0
3 years ago
Read 2 more answers
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