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2 years ago

Why would 60g at 10C of NaCl in 200g of water would be unsaturated?

Chemistry

1 answer:

Volgvan2 years ago

5 0

Answer:

The amount of substance needed to saturate water increases as the temperature increases.

Explanation:

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Question 6 of 10

love history [14]

B. Newton’s second law

6 0

2 years ago

A backpacker carries 2.7 L of white gas as fuel for her stove.How many pounds does the fuel add to her load? Assume the density

Zepler [3.9K]

2.7 L in cm³ :

2.7 * 1000 = 2700 cm³

Weight = Volume * Density

2700 * 0.79 = 2133 g

1 Ibs = <span>453.59 g

2133 / 453.59 = 4.70 Ibs

hope this helps!</span>

6 0

2 years ago

What is the pH of a 0.222 M acetic acid solution?

Oksi-84 [34.3K]

pH=2.7

<h3>Further explanation</h3>

Acetic acid = weak acid

$\tt [H^+]=\sqrt{Ka.M}$

Ka = acid ionization constant

M = molarity

Ka for Acetic acid(CH₃COOH) : 1.8 x 10⁻⁵

$\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.222}\\\\=0.001998=1.998\times 10^{-3}$

$\tt pH=3-log~1.998=2.7$

7 0

2 years ago

Calculate the pH of 1.00 X 10^-6 M HCI *

murzikaleks [220]

Answer: 6

Explanation:

To find pH you have to do -log(concentration of H+)

7 0

2 years ago

Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)

allsm [11]

Answer:

$\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}$

Explanation:

1. Density from mass and volume

$\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}$

2. Volume from density and mass

$V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}$

3. Mass from density and volume

$\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}$

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water =<u> 12.8 mL</u>

Volume of object = 11.8 mL

$\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}$

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

6 0

3 years ago