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3 years ago

A student completely dissolves some salt in a beaker of water. Which of these statements does NOT describe salt and water?

Chemistry

1 answer:

aleksklad [387]3 years ago

3 0

Answer:

It is a compound

Explanation:

I got the answer right on the quiz.

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A 500.0 mL sample of 0.18 MHClO4 is titrated with 0.27 MCsOH. Determine the pH of the solution after the addition of 250.0 mL of

NNADVOKAT [17]

Answer:

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Explanation:

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2 years ago

Calculate the number of moles present in 9.50g of co2

pshichka [43]

Answer:

0.4318 mol

Explanation:

Moles= mass/ mollar mass

Moles= 9.50/22

Moles= 0.4318 mol

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3 years ago

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3 years ago

Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as molarity, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

n = moles (unit mol)

V = volume (unit L)

Find the molar mass (M) of potassium hydroxide.

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

Calculate the moles of potassium hydroxide.

$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

Convert the volume of water to litres.

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

Calculate the concentration.

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0

2 years ago

What is created when an acid is mixed with a base?

Arturiano [62]

Answer:

If you mix equal amounts of a strong acid and a strong base, the two chemicals essentially cancel each other out and produce a salt and water. Mixing equal amounts of a strong acid with a strong base also produces a neutral pH (pH = 7) solution.

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2 years ago