What is the process that changes the composition of rocks by dissolving them called?

Chemistry

2 answers:

Hoochie [10]3 years ago

5 0

Chemical weathering
..................................

AlladinOne [14]3 years ago

3 0

Weathering & Erosion

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Hurry I need help quickThe table shows the mass and boiling point of some substances. Boiling Point of Substances Mass (kg) Boil

Talja [164]

I am probably wrong but I would think 900. I wouldn’t submit that without somebody backing me up though.

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4 years ago

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A sealed container of nitrogen gas is at 1000.0 kPa pressure and at a temperature of 20.00 °C. The container is left in the sun,

vodka [1.7K]

Answer:

P₂ = 1102 Kpa

Explanation:

Given data:

Initial pressure = 1000.0 Kpa

Initial temperature = 20.0°C

Final temperature = 50.0°C

Final pressure = ?

Solution:

Initial temperature = 20.0°C (20+273 = 293 K)

Final temperature = 50.0°C (50+273 = 323 K)

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

1000 Kpa / 293 K = P₂/323 K

P₂ = 1000 Kpa × 323 K / 293 K

P₂ = 323000 Kpa. K /293 K

P₂ = 1102 Kpa

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3 years ago

Fluoride ion is poisonous in relatively low amounts: 0.2 g of F- per 70 kg of body weight can cause death. Nevertheless, in orde

svetoff [14.1K]

Answer:

$\boxed{\text{(a) 14 L; (b) 711 kg}}$

Explanation:

(a) Litres of water

$\text{Mass of F}^{-} = 8.5 \times 10^{7}\text{ gal} \times \dfrac{\text{0.2 mg F}^{-}}{\text{1 kg}} = \textbf{14 mg F}^{-}\\\\\text{Volume of water} = \text{14 mg F}^{-} \times \dfrac{\text{1 L water}}{\text{1 mg F}^{-}} = \textbf{14 L water}\\\\\text{The person would have to consume $\boxed{\textbf{14 L}}$ of water per day}$

(b) Mass of NaF

$\text{Volume} =8.5 \times 10^{7} \text{gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} = 3.22 \times 10^{8}\text{ L}\\\\\text{Milligrams of F}^{-} = 3.22 \times 10^{8}\text{ L} \times \dfrac{\text{ 1 mg F}^{-}}{\text{1 L}} = 3.22 \times 10^{8}\text{ mg F}^{-}\\\\\text{kilograms of F}^{-} = 3.22 \times 10^{8}\text{ mg F}^{-} \times \dfrac{\text{1 mg F}^{-}}{10^{6}\text{kg F⁻}} = \text{322 kg F}^{-}$

1 kmol of NaF (41.99 kg) contains 19.00 kg of F⁻.

$\text{Kilograms of NaF}= \text{322 kg F}^{-} \times \dfrac{\text{41.99 kg NaF}}{\text{19.00 kg F}^{-}} = \text{711 kg NaF}\\\\\text{It will take } \boxed{\textbf{711 kg}} \text{ of NaF to treat the reservoir}$