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Romashka-Z-Leto [24]
3 years ago
14

a balloon will get smaller​ when it's filled with air, when air is heated, or cooled, or when it rises?

Chemistry
2 answers:
Irina-Kira [14]3 years ago
8 0
As it air is cooled, a balloon will get smaller. 

to better understand this, we should recall the properties of gases

when we have pressure, volume and temperature, they are all proportional to each except pressure and volume. 

so, if air is heated, meaning temperature is increased, that means volume increases. (get's bigger-- not the answer)

if balloon rises, as it goes up, the pressure decreases, if pressure decreases, volume increases

when the air is cooled, meaning temperature is decreased, that means volume decreases as well. 
lora16 [44]3 years ago
5 0
Assuming the balloon is untied aka the air inside isn't intact.

It will get smaller when it rises.
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xrays are used to diagnose diseases of internal body organs. What is the frequency of an X ray with a wavelentgh of 1.15*10^-10
Reil [10]

Answer:

2.31 EHz

Explanation:

The formula relating the frequency (f) and wavelength (λ) is

fλ = c                                        Divide both sides by λ

f = c/λ

c = 2.998 × 10⁸ m·s⁻¹

λ = 1.15 × 10⁻¹⁰ m                     Calculate f.

f = 2.998 × 10⁸/1.15 × 10⁻¹⁰

f = 2.61 × 10¹⁸ s⁻¹ = 2.61 EHz

5 0
3 years ago
Read 2 more answers
Can someone help me with this please it's for friday? We are giving the VSEPR theory. I give 100 points.​
Norma-Jean [14]

Answer: Sulfite ion has a shorter

S-O

bonds.

Explanation:

The Lewis structure of

SO

3

is

SO3

SO3

(from chemistry.stackexchange.com)

It has a total of six σ and π bonds to the three

O

atoms.

The average bond order of an

S-O

bond is

6

3

=

2

.

The Lewis structure of

SO

2

−

3

is

It is a resonance hybrid, with a total of four σ and π bonds to the three

O

atoms.

The average bond order of an

S-O

bond is

4

3

≈

1.33

.

Since bond length decreases as bond order increases,

SO

3

is predicted to have the shorter bond length.

Sulfite

Sulfite

Explanation:

3 0
3 years ago
A solution that is 0.135 M is diluted to make 500.0 mL of a 0.0851 M solution. How many milliliters of the original solution wer
sineoko [7]

Answer:

315mL

Explanation:

Data obtained from the question include the following:

Molarity of stock solution (M1) = 0.135 M

Volume of stock solution needed (V1) =?

Molarity of diluted solution (M2) = 0.0851 M

Volume of diluted solution (V2) = 500mL

The volume of the stock solution needed can be obtain as follow:

M1V1 = M2V2

0.135 x V1 = 0.0851 x 500

Divide both side by 0.135

V1 = (0.0851 x 500) / 0.135

V1 = 315mL

Therefore, the volume of the stock solution needed is 315mL

6 0
3 years ago
A sample of gas at 10.0 ATM and 5.0 °C increases in tempature of 35 °C. If the volume is unchanged, what is the new pressure?
agasfer [191]

Answer:

11.08 atm

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 10 atm

Initial temperature (T₁) = 5 °C

Final temperature (T₂) = 35 °C

Final pressure (P₂) =?

Volume = constant

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 5 °C

Initial temperature (T₁) = 5 °C + 273

Initial temperature (T₁) = 278 K

Final temperature (T₂) = 35 °C

Final temperature (T₂) = 35 °C + 273

Final temperature (T₂) = 308 K

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

Initial pressure (P₁) = 10 atm

Initial temperature (T₁) = 278 K

Final temperature (T₂) = 308 K

Final pressure (P₂) =?

P₁/T₁ = P₂/T₂

10/278 = P₂/308

Cross multiply

278 × P₂ = 10 × 308

278 × P₂ = 3080

Divide both side by 278

P₂ = 3080 / 278

P₂ = 11.08 atm

Therefore, the final pressure of the gas is 11.08 atm

3 0
3 years ago
How much heat (in kJ) would need to be removed to cool 150.3 g of water from 25.60°C to -10.70°C?
olga_2 [115]

Answer:

Q = -22.9 kJ

Explanation:

Given that,

Mass of water, m = 150.3 g

Water gets cool from 25.60°C to -10.70°C.

The specific heat of water, c = 4.2 J/g°C

The formula for heat needed is given by :

Q=mc\Delta T\\\\Q=150.3\times 4.2 \times (-10.7-25.6)\\\\Q=-22914.738\\\\or\\\\Q=22.9\ kJ

So, 22.9 kJ of heat is needed to be removed to cool.

8 0
3 years ago
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