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ZanzabumX [31]
3 years ago
12

Describe the relationship between gravitational force and distance as shown in the diagram.

Chemistry
1 answer:
aev [14]3 years ago
3 0

Answer:

as you move away from a planet the gravitational force gets weaker. just think, if you get close towards a sun then the gravitational force will be stronger.

You might be interested in
PLEASE HELP...............................................
Serjik [45]
Umm its not a i think... lemme see lol

5 0
3 years ago
PLEASE HELP!!
grandymaker [24]

Answer:

1. 136 °C.

2. 0.21 atm.

Explanation:

1. Determination of the new temperature in °C.

Initial volume (V1) = 1.35L

Final volume (V2) = 1.95L

Initial temperature (T1) = 283 K

Final temperature (T2) =...?

Using the Charles' law equation, the new temperature of the gas can be obtained as follow:

V1 /T1 = V2 /T2

1.35/283 = 1.95/T2

Cross multiply

1.35 × T2 = 283 × 1.95

1.35 × T2 = 551.85

Divide both side by 1.35

T2 = 551.85/1.35

T2 = 408.8 ≈ 409 K

Finally, we shall convert 409 K to °C. This can be obtained as follow:

T (°C) = T(K) – 273

T(K) = 409 K

T (°C) = 409 – 273

T (°C) = 136 °C

Therefore, the new temperature of the gas is 136 °C.

2. Determination of the new pressure.

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1.67 L

Final pressure (P2) =.?

Next, we shall convert 1.67 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

1.67 L = 1.67 L × 1000 mL / 1 L

1.67 L = 1670 mL

Therefore, 1.67 L is equivalent to 1670 mL.

Finally, we shall determine the new pressure of the gas as follow:

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1670 mL

Final pressure (P2) =.?

P1V1 = P2V2

1.34 × 267 = P2 × 1670

357.78 = P2 × 1670

Divide both side by 1670.

P2 = 357.78 / 1670

P2 = 0.21 atm.

Therefore, the new pressure of the gas is 0.21 atm.

3 0
3 years ago
.
nydimaria [60]
Is the answer Radiation
4 0
2 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
2 years ago
Please, Give the correct answer ! Don't spam comment​
mars1129 [50]

Answer:

hxuyufydufiduud

Explanation:

chdckjxxhtxgx

5 0
3 years ago
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