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3 years ago

Place the following substances in the following order: soluble, somewhat soluble, and insoluble.

1 answer:

8 0

C- Sugar is very much soluble (example- putting sugar in tea), sand is definately not soluble (example- a sandy beach) and cornstarch is the last one, so logically it should be somewhat soluble.

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A chemist prepares a solution of sodium thiosulfate by measuring out of sodium thiosulfate into a volumetric flask and filling t

marshall27 [118]

Mass of sodium thiosulfate $Na_{2}S_{2}O_{3}$ is 110. g

Volume of the solution is 350. mL

Calculating the moles of sodium thiosulfate:

$110. g Na_{2}S_{2}O_{3} * \frac{1 mol Na_{2}S_{2}O_{3}}{158.1 g Na_{2}S_{2}O_{3}}$ = 0.696 mol $Na_{2}S_{2}O_{3}$

Converting the volume of solution to L:

$350. mL * \frac{1 L}{1000 mL} = 0.350 L$

Finding out the concentration of solution in molarity:

$\frac{0.696 mol}{0.350 L} = 1.99 mol/L$

4 0

3 years ago

What is the ratio of hydrogen atoms to sulfur atoms in sulfuric acid, h2so4?

Vsevolod [243]

I think it's 2:1 or 2:1:4. I mostly think it's 2:1 though. (:

4 0

3 years ago

Given the reaction _K(s) +_ Cl2(g) → _KCl(s) what is the amount of K, in grams, needed to completely react with 2 moles of Cl2(g

damaskus [11]

Answer:

156.4g K

Explanation:

I'm not sure if it is correct but I think it should be this

What do we know so far?: 2K + 1Cl2 -> 2KCl, 2 mol of Cl2

What are we looking for?: #g of K

What is the ratio of K to Cl2?: 2:1

Set up equation: 2molCl2 x $\frac{2mol K}{1 mol Cl2}$

Cancel unwanted units: 2 x $\frac{2mol K}{1}$

Answer we got: 2 x 2mol K = 4mol K

Converting moles to grams: 4 x 39.1 (molar mass of K) = 156.4g K

3 0

2 years ago

Consider the nuclear equation below.

Dovator [93]

Answer:

its NOT C

Explanation:

8 0

3 years ago

Read 2 more answers

A 100.0mL bubble of hot gases at 225 C and 1.80 atm escapes from an active volcano, what is the new volume of the bubble outside

Inessa05 [86]

<h3>Answer:</h3>

112.08 mL

<h3>Explanation:</h3>

From the question we are given;

Initial volume, V1 = 100.0 mL
Initial temperature, T1 = 225°C, but K = °C + 273.15

thus, T1 = 498.15 K

Initial pressure, P1 = 1.80 atm
Final temperature , T2 = -25°C

= 248.15 K

Final pressure, P2 = 0.80 atm

We are required to calculate the new volume of the gases;

According to the combined gas law equation;

$\frac{P1V1}{T1}=\frac{P2V2}{T2}$

Rearranging the formula;

$V2=\frac{P1V1T2}{T1P2}$

Therefore;

$V2=\frac{(1.80atm)(100mL)(248.15K)}{(498.15K)(0.80atm)}$

$V2=112.08mL$

Therefore, the new volume of the gas is 112.08 mL

8 0

3 years ago