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2 years ago

One mole of water weights?

Chemistry

1 answer:

Gemiola [76]2 years ago

8 0

Answer:

c 18.0ml

Explanation:

The average mass of one H2O molecule is 18.02 amu. The number of atoms is an exact number, the number of mole is an exact number; they do not affect the number of significant figures. The average mass of one mole of H2O is 18.02 grams. This is stated: the molar mass of water is 18.02 g/mol.

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Imagine a friend is planning a rock-climbing trip. Write a note in the box explaining how gas exchange is affected at the top of

wariber [46]

Answer:

Air is less dense on a mountaintop than at sea level.

Air pressure is lower at low altitudes.

As you climb a mountain, air pressure increases.

More force pushes on the air at the bottom of an air column.

As you descend a mountain, air molecules are closer together.

Explanation:

6 0

2 years ago

K-12 Molar Mass 7/10 Calculate the molar mass of aluminum nitrate, Al(NO3)3. g/mol

tigry1 [53]

Molar mass = 212.996 g/mol

,,

You basically add the total atomic weight, these weights can be found on a periodic table

4 0

3 years ago

If a sample of HF gas at 694.9 mmHg has a volume of 3.463 Land the volume is changed to 5.887 L, then what will be the new press

VARVARA [1.3K]

Answer:

$\large \boxed{\text{381.7 mmHg}}$

Explanation:

Data:

p₁ = 694.9 mmHg; V₁ = 3.463 L

p₂ = ?; V₂ = 5.887 L

Calculation:

$\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{648.9 mmHg} \times \text{3.463 L} & = & p_{2} \times\text{5.887 L}\\\text{2247.1 mmHg} & = & 5.887p_{2}\\p_{2} & = & \dfrac{\text{2247.1 mmHg}}{5.887}\\\\& = &\textbf{381.7 mmHg}\\\end{array}\\\text{The new pressure of the gas is $\large \boxed{\textbf{381.7 mmHg}}$}$

4 0

3 years ago

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0

3 years ago

A solution is known to contain only one type of cation. Addition of Cl1- ion to the solution had no apparent effect, but additio

zhannawk [14.2K]

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate $SO_4~^-^2$ and the chloride $Cl^-$ :

Sulfate

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.( $Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2$ ), which are NOT soluble.

Chloride

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ( $Pb^+^2~,Ag^+~,Hg_2~^+^2$ ), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I), lead" and our possibilities are:

"Calcium, strontium, or barium".

I hope it helps!

7 0

3 years ago