All categories

3 years ago

Name two unique properties of water

Chemistry

1 answer:

arsen [322]3 years ago

7 0

Water's high polarity
Water's attraction to other polar molecules
I hope this works!!

You might be interested in

An object at a distance of 30 cm from a concave mirror gets its image at the same point. The focal length of the mirror is

Nimfa-mama [501]

The focal length of the mirror is 15 cm

From the question given above, the following data were obtained:

Object distance (u) = 30 cm

Image distance (v) = 30 cm

<h3>Focal length (f) =? </h3>

The focal length of the mirror can be obtained as follow:

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\\\\frac{1}{f} = \frac{1}{30} + \frac{1}{30}\\\\\frac{1}{f} = \frac{2}{30} \\\\\frac{1}{f} = \frac{1}{15} \\\\$

<h3>Invert </h3><h3>f = 15 cm</h3>

Therefore, the focal length of the mirror is 15 cm

Learn more: brainly.com/question/1392083

3 0

3 years ago

Show

SIZIF [17.4K]

Answer: 0.9375 g

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$ .....(1)

Molarity of $HCl$ solution = 0.75 M

Volume of $HCl$ solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Moles of} HCl={0.75}\times{0.025}=0.01875moles$

$CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(s)+CO_2(g)+H_2O(l)$

According to stoichiometry :

2 moles of $HCl$ require = 1 mole of $CaCO_3$

Thus 0.01875 moles of $HCl$ will require= $\frac{1}{2}\times 0.01875=0.009375moles$ of $CaCO_3$

Mass of $CaCO_3=moles\times {\text {Molar mass}}=0.009375moles\times 100g/mol=0.9375g$

Thus 0.9375 g of $CaCO_3$ is required to react with 25.0 ml of 0.75 M HCl

6 0

3 years ago

If the sun rises at 7:36am and sets at 6:00pm, what times would we use for our data in the 24-hour clock format?

Dmitrij [34]

C is the answer I used to work at McDonald’s I know military time

6 0

3 years ago

Good Morning I need help with this science work, please help it would mean a lot.

Kay [80]

That’s easy it would be B.

8 0

3 years ago

Calculate the molarity when 135 g of KCl is dissolved to make a 2.50 L solution. Round to two significant digits.

inysia [295]

Explanation:

$\textrm{Molarity of a solution}=\frac{\textrm{Number of moles of solvent}}{\textrm{Volume of solution in Liters}}$

We are given a $2.50L$ solution which contains $135g$ of $KCl$ .

To calculate number of moles of $KCl$ present, we need to find it's molar mass from charts. Molar Mass of $KCl$ is known to be $74.55\frac{g}{Mol}$ .

Number of moles of $KCl$ present = $\frac{\textrm{Given weight}}{\textrm{Molar Mass}}\textrm{ = }\frac{135g}{74.55\frac{g}{Mol}}\textrm{ = }1.81087mol$

Molarity = $\frac{1.81087mol}{2.50L}=0.7243\frac{mol}{L}$

∴ Molarity of given $KCl$ solution = $0.72\frac{mol}{L}$

7 0

3 years ago