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2 years ago

2.37 x 0.004 Sig Fig

Chemistry

2 answers:

8090 [49]2 years ago

8 0

Answer:

0.00948

Explanation:

NeTakaya2 years ago

5 0

Answer:

The answer is 0.00948

Explanation:

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The is the si unit that expresses the amount of substance. Specifically, it is defined as the amount of substance containing the

Dmitry_Shevchenko [17]

The Mole is the SI unit that expresses the amount of substance.

Mole is defined as - The mole is the amount of substance containing the same number of entities as there are in the 12 grams of Carbon - 12.

Mole is denoted by using symbol mol.

Mole = 6.022 x 10²³ elementary entities.

These number of elementary entities in 1 mole is equal to or called as an Avogadro's number. Mole is equal to 6.022 x 10²³ because this number of entity is same as in exactly 12 g of carbon-12.

It is a very important SI unit of measured which is used by the chemists. Moles are used in measuring in small or tiny things such as atoms, molecules and the other tiny particles.

To learn more about the mole concept,

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3 0

1 year ago

Sulfur hexafluoride, a dense gas, is held in two separate containers in a storage room at an atmospheric pressure of 755 mmHg an

dimaraw [331]

Explanation:

As it is given that both the given containers are at same temperature and pressure, therefore they have the same density.

So, mass of $SF_{6}$ in container- 1 is as follows.

5.35 mol x molar mass of $SF_{6}$

= 7.61 mol x 146.06 g/mol

= 1111.52 g

Therefore, density of $SF_{6}$ will be calculated as follows.

Density = $\frac{mass}{volume}$

density = $\frac{1111.52 g}{2.09 L \times 1000 ml/L}$

= 0.532 g/mL

Now, mass of $SF_{6}$ in container- 2 is calculated as follows.

4.46 L x 1000 mL/L x 0.532 g/mL

= 2372.72 g

Hence, calculate the moles of moles $SF_{6}$ present in container 2 as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{2372.72 g}{146.06 g/mol}$

= 16.24 mol

Since, 1 mol $SF_{6}$ contains 6 moles F atoms .

So, 16.24 mol $SF_{6}$ contains following number of atoms.

= $16.24 mol \times 6$

= 97.46 mol

Thus, we can conclude that moles of F atoms in container 2 are 97.46 mol.

7 0

3 years ago

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution before the addition of any HNO3.

alex41 [277]

Answer:

$pH=11.12$

Explanation:

Hello,

In this case, ammonia dissociation is:

$NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)$

So the equilibrium expression:

$Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}$

That in terms of the reaction extent and the initial concentration of ammonia is written as:

$1.8x10^{-5}=\frac{x*x}{0.10M-x}$

Thus, solving by using solver or quadratic equation we find:

$x=0.00133M$

Which actually equals the concentration of hydroxyl ion, therefore the pOH is computed:

$pOH=-log([OH^-])=-log(0.00133)=2.88$

And the pH from the pOH is:

$pH=14-pOH=14-2.88\\\\pH=11.12$

Best regards.

6 0

3 years ago

For the set of ionic compounds, LiCl, AgCl, PbCl2, choose the correct characterization of their solubilities in water from the r

sergiy2304 [10]

Answer:

Two of the three salts are soluble.

One of the three salts is soluble.

Explanation:

One or two of the three salts would be soluble in water depending on if the water is cold or hot.

According to the solubility rules of chloride salts, all chlorides are soluble in cold water except for those of silver (Ag), lead (Pb), and mercury (I) (Hg).

This thus means that both AgCl and $PbCl_2$ would be insoluble in cold water while LiCl would be soluble.

$PbCl_2$ is, however soluble in hot water.

Hence, in cold water, one of the three salts (LiCl) would be soluble, while in hot water, two of the three salts (LiCl and $PbCl_2$ ) would be soluble.

7 0

3 years ago

What is the elevation in boiling point of a 2.39 molal aq. solution of glucose?

Karo-lina-s [1.5K]

1. What are the answers? and 2. It's b.

6 0

3 years ago