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julia-pushkina [17]
2 years ago
11

A 29.0-g sample of water at 290. K is mixed with 51.0 g water at 300. K. Calculate the final temperature of the mixture assuming

no heat loss to the surroundings.
Chemistry
2 answers:
zubka84 [21]2 years ago
5 0
Let me conduct my research
Anvisha [2.4K]2 years ago
3 0

answer: 296 K

steps:

mass times temperature =mass times temperature

-51 * 300 = 29 * -290

300 * 51 = 15300

290 * 29 = 8410

-51x + 15300 = 29x - 8410

51 + 29 = 80

15300 + 8410 = 23710

23710 = 80x

80x = 23710

x = 296.375

https://socratic.org/questions/a-30-0-g-sample-of-water-at-280-k-is-mixed-with-50-0-g-of-water-at-330-k-how-wou

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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
5. Write the names of the following ions.
Talja [164]
A.) Phosphate ion or Orthophosphate
d.) Hydroxide
D.) Ammonium
e.) Iron
C.) Nitrate
f.) Sulfur dioxide
5 0
2 years ago
Be sure to answer all parts.
Marianna [84]

a)3^2 = 9 times

b)2^2*2 = 8 times

c)(1/2) = half times

7 0
3 years ago
MgCl2 + AgNO3 → AgCl + Mg(NO3)2
Law Incorporation [45]

Answer:

MgCl2 + 2AgNO3 → 2AgCl + Mg(NO3)2

Explanation:

I'm assuming you want to balance it so...

The first thing I see is that there are two chlorines on the reactant side and one on the product side

Adding a coefficient of 2 would get 2AgCl2

Now there are two silvers on the reactant side, so add a 2 to AgNO3 on the products side. Now they are all balanced.

If that is not what you are looking for let me know!

6 0
3 years ago
Read 2 more answers
Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:(a) |(b) Sr(c) K(d) N(e) S(f)
Pavel [41]

Answer:

(a) I⁻ (charge 1-)

(b) Sr²⁺ (charge 2+)

(c) K⁺ (charge 1+)

(d) N³⁻ (charge 3-)

(e) S²⁻ (charge 2-)

(f) In³⁺ (charge 3+)

Explanation:

To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.

(a) |

I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).

(b) Sr

Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).

(c) K

K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).

(d) N

N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).

(e) S

S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).

(f) In

In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).

3 0
3 years ago
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