When elements that form a mineral dissolve in hot water, they form a mixture called a(n) ______.?

Chemistry

1 answer:

Serhud [2]3 years ago

4 0

Soluble materials when put in a hot water would dissolve.

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Match each word to it's correct meaning.

kvasek [131]

1 is The first one
2 is the third one
3 is the second one
4 is the 4th one

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2 years ago

A solution is made by dissolving 23.0 g KOH in 1.60 L H2O. What is the molality of the solution? The density of water is 1.00 g/

artcher [175]

The molarity of a solution is defined as the moles of solute dissolved per kilogram of solvent. Therefore, we first compute the moles of KOH using:

Moles = mass / Mr
Moles = 23 / 56
Moles = 0.41

The volume of solvent is 1.6 liters
The density is 1 gram/cm³ = 1 kg/L
Mass of solvent = density * volume
Mass of solvent = 1 * 1.6
Mass of solvent = 1.6 kg

Molality = moles / kilogram
Molarity = 0.41 / 1.6
Molarity = 0.26

The molality of the solution is 0.26 molal.

4 0

3 years ago

Read 2 more answers

When an atom becomes an anion it becomes Smaller,Larger,neutral,positive

liraira [26]

When an atom becomes an anion it loses ions, so its becoming smaller

3 0

2 years ago

When the volume of a gas is

Montano1993 [528]

Answer:

$\boxed {\boxed {\sf 232.9 \textdegree C}}$

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

$\frac {V_1}{T_1}= \frac{V_2}{T_2}$

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

$\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}$

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

$\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

$250 \ mL * T_2 = 137 \textdegree C * 425 \ mL$