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3 years ago

Complete the statements by writing the number

Chemistry

1 answer:

dezoksy [38]3 years ago

3 0

Answer :

The substance is in the gas phase only in region → 5

The substance is in both the liquid and the solid phase in region → 2

The substance is in only the liquid phase in region → 3

The melting point is the temperature at region → 2

The boiling point is the temperature at region → 4

Explanation :

Six phases of substance:

Melting or fusion : In this process the phase changes from solid state to liquid state at constant temperature.

Freezing : In this process the phase changes from liquid state to solid state at constant temperature.

Evaporation : In this process the phase changes from liquid state to gaseous state at constant temperature.

Condensation : In this process the phase changes from gaseous state to liquid state at constant temperature.

Sublimation : In this process the phase changes from solid state to gaseous state without passing through the liquid state at constant temperature.

Deposition : In this process the phase changes from gaseous state to solid state without passing through the liquid state at constant temperature.

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How many moles of Kr are contained in 398 mg of Kr?

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The mass of 1 mole of an element is its atomic weight on the Periodic Table in grams.

1g = 1000mg

The mass of 1 mole of Kr = 83.80g Kr

Convert mg Kr to g Kr.

398mg Kr x (1g/1000mg) = 0.398g Kr

Convert g Kr to mol Kr.

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4 years ago

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3 years ago

One of relatively few reactions that takes place directly between two solids at room temperature is Ba(OH)2 · 8H2O + ammonium th

Hatshy [7]

Answer:

$\boxed{\text{3.1 g}}$

Explanation:

We will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 315.46 76.12

Ba(OH)₂·8H₂O + 2NH₄SCN ⟶ Ba(SCN)₂ + 2NH₃ + 10H₂O

m/g: 6.5

1. Moles of Ba(OH)₂·8H₂O

$\text{Moles of Ba(OH)$_{2}\cdot $8H$_{2}$O}\\= \text{ 6.5 g Ba(OH)$_{2}\cdot $8H$_{2}$O} \times \dfrac{\text{1 mol Ba(OH)$_{2}\cdot $8H$_{2}$O}}{\text{ 315.46 g Ba(OH)$_{2}\cdot $8H$_{2}$O}}\\= \text{0.0206 mol Ba(OH)$_{2}\cdot $8H$_{2}$O}$

2. Moles of NH₄SCN

The molar ratio is 2 mol NH₄SCN:1 mol Ba(OH)₂·8H₂O

$\text{Moles of NH$_{4}$SCN} =\text{0.0206 mol Ba(OH)$_{2}\cdot $8H$_{2}$O} \times \dfrac{\text{2 mol NH$_{4}$SCN}}{\text{1 mol Ba(OH)$_{2}\cdot $8H$_{2}$O}}\\= \text{0.0412 mol NH$_{4}$SCN}$

3. Mass of NH₄SCN

$\text{Mass of NH$_{4}$SCN} = \text{0.0412 mol NH$_{4}$SCN} \times \dfrac{\text{76.12 g NH$_{4}$SCN}}{\text{1 mol NH$_{4}$SCN}} =\\\textbf{3.1 g NH$_{4}$SCN}\\\\\text{You must use }\boxed{\textbf{3.1 g}}\text{ NH$_{4}$SCN}$