We do we mean when we say a substance is pure

Chemistry

1 answer:

tiny-mole [99]3 years ago

6 0

Answer:

ot means that contains only one kind of matter.

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Which of the following are the physical properties of compounds? Select all that apply.

BlackZzzverrR [31]

A. Smell (Oder) B. Color and D I’m pretty sure because the common properties are Color, Oder, Melting point, and boiling point

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3 years ago

Phosphorus has an atomic mass of 31 and an atomic number of 15, so the number of neutrons must be _______________.

Nikitich [7]

Remember this:

atomic mass= atomic number + number of neutrons

The atomic number (same thing as number of protons) in this case is 15. The atomic mass is 31.

So, we subtract 15 from 31 to find the number of neutrons.

31-15=16

16 neutrons!!!

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3 years ago

An atom that has 13 protons and 15 neutrons is isotope of the element

wariber [46]

Answer=c

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3 years ago

What are giant covalent structures

PIT_PIT [208]

Covalent bonding - giant covalent structures. ... The atoms are usually arranged into giant regular lattices - extremely strong structures because of the many bonds involved. The graphic shows the molecular structure of diamond and graphite: two allotropes of carbon, and of silica.

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3 years ago

Suppose that 98.0g of a non electrolyte is dissolved in 1.00kg of water. The freezing point of this solution is found to be -0.4

Sholpan [36]

Answer:

$\large \boxed{\text{392 u}}$

Explanation:

1. Calculate the molal concentration

The formula for the freezing point depression by a nonelectrolyte is

$\Delta T_{f} = -K_{f}b\\b = -\dfrac{\Delta T_{f}}{ K_{f}} = -\dfrac{-0.465 \, ^{\circ}\text{C}}{\text{1.86 $\, ^{\circ}$C$\cdot$kg$\cdot$mol}^{-1}} = \text{0.250 mol/kg}$

2. Calculate the moles of solute

$\begin{array}{rcl}b & = & \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}\\\\\text{moles of solute} & = & b \times {\text{kilograms of solvent}}\\n & = &\text{0.250 mol/kg} \times \text{1.00 kg}\\ & = & \text{0.250 mol}\\\end{array}$

3. Calculate the molecular mass

$\begin{array}{rcl}\text{Moles} & = &\dfrac{\text{mass}}{\text{molar mass}}\\\\\text{0.250 mol} & = & \dfrac{\text{98.0 g}}{MM}\\\\MM & = & \dfrac{\text{98.0 g }}{\text{0.250 mol}}\\\\& = &\textbf{392 g/mol}\\\end{array}\\\text{The molecular mass of the solute is $\large \boxed{\textbf{392 u}}$}$

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4 years ago