All categories

4 years ago

Does adding salt to water lower its freezing point science project

1 answer:

8 0

Yes, it does. The freezing point of the salt solution is lower than the freezing point of pure water since there will be more particles, which are water molecules plus salt molecules, present in the salt solution. The salt molecules will slow down the amount of crystals formed resulting to a lower freezing point.

You might be interested in

which of the following solution is a buffer? a solution containing Hf and NaOH, NH3 and NH4+, HCN and HNO3?

Ivahew [28]

The answer is NH3 an NH4+. Buffer solutions contain a weak base and its conjugate acid or a weak acid and its conjugate base. NH3 is a weak base and NH4+ is its conjugate acid.

6 0

3 years ago

What caused wind to blow

Arlecino [84]

Answer:

Different atmospheric pressure. When there is a different atmospheric pressure, air moves from the higher pressure to the lower pressure area which results in what you call <u>WIND</u> but can result in various speeds and pressure.

Hope this helped and if it did, please give my answer a brainliest.

7 0

4 years ago

Read 2 more answers

If 0.2 M Ag+, 2 M Ba2+, and 2 M Ca2+ are added to a Na2SO4 solution, which will precipitate first? (Ksp of Ag2SO4 = 1.4 x 10-5,

Sedbober [7]

A.

This is because BaSO₄ has the lowest solubility product out of all the substances.

7 0

3 years ago

How many moles of potassium sulfate are present in a sample that weighs 45.8g?

Nikitich [7]

Answer:

0.263 mol

Explanation:

K2SO4 - potassium sulfate

Molar mass (K2SO4) = 2*39.0983 + 32.065 + 4*15.9994 =

= 174.2592≈174.26 g/mol

45.8 g * 1 mol/174.26 g = 0.263 mol

5 0

3 years ago

A pool is 60.0 m long and 35.0 m wide. How many mL of water are needed to fill the pool to an average depth of 6.35 ft? Enter yo

ra1l [238]

Answer:

4.07 × 10⁹ mL

Explanation:

Step 1: Given data

Length of the pool (L): 60.0 m

Width of the pool (W): 35.0 m

Depth of water in the pool (D): 6.35 ft

Step 2: Convert "D" to m

We will use the relationship 1 m = 3.28 ft

$6.35ft \times \frac{1m}{3.28ft} = 1.94m$

Step 3: Calculate the volume of water (V)

We will use the following expression.

$V = L \times W \times D = 60.0m \times 35.0m \times 1.94 m = 4.07 \times 10^{3} m^{3}$

Step 4: Convert "V" to mL

We will use the relationship 1 m³ = 10⁶ mL.

$4.07 \times 10^{3} m^{3} \times \frac{10^{6}mL }{1m^{3} } = 4.07 \times 10^{9} mL$

7 0

3 years ago