Circulatory Quiz - 1 chance

Hydrogen gas is collected over water in an inverted buret. If the atmospheric pressure is 745 mm Hg, the vapor pressure of water

Varvara68 [4.7K]

Answer:

$P_{H_2}=727mmHg$

Explanation:

Hello there!

In this case, according to the given data, it is possible to infer that the gas mixture lies on the 15.0 cm-high column of water, so that the total pressure or atmospheric pressure is given by:

$P_{atm}=P_{water}+P_{H_2}$

Thus, since the atmospheric pressure is 745 mmHg and the vapor pressure of water is 18 mmHg, the pressure of hydrogen turns out to be:

$P_{H_2}=P_{atm}-P_{water}\\\\P_{H_2}=745mmHg-18mmHg\\\\P_{H_2}=727mmHg$

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6 0

2 years ago

A 2.026g sample of a hydrate of sodium carbonate (Na2CO3) is heated to remove water. After heating, the mass of the sample is 0.

OLEGan [10]

Answer:

62.98 % of the sample of hydrate is water

Explanation:

Step 1: Data given

Mass of the sample of a hydrate of sodium carbonate (Na2CO3) = 2.026 grams

After heating, the mass of the sample is 0.750 g

Molar mass H2O = 18.02 g/mol

Step 2: Calculate mass of water

Mass water = mass of hydrate - mass of sample after heating

Mass water = 2.026 grams - 0.750 grams

Mass water = 1.276 grams

Step 3: Calculate mass % percent of water

Mass % of water = (mass of water / total mass hydrate) * 100 %

Mass % of water = (1.276 grams / 2.026 grams) *100 %

Mass % of water = 62.98 %

62.98 % of the sample of hydrate is water

7 0

2 years ago

Read 2 more answers

given the following reaction: CuO (s) + H2 (g) to Cu(s) + H2O (g). If 250 L of hydrogen gas are used to reduce copper (II) oxide

QveST [7]

24g is needed no explanation

4 0

2 years ago

SeF6 1. Lewis Structure 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs

Anarel [89]

The answer is- $SF_{6}$ is octahedral in electronic and molecular geometry with 6 Fluorine atoms bonded to central atom S.

Lewis structures are the diagrams in which the valence electrons of the atoms of a compound are arranged around the atoms showing the bonding between the atom and the lone pair of electrons existing in the molecule.

Determine the molecular geometry of $SF_{6}$ .

Valence Shell Electron Pair Repulsion theory is commonly known as VSEPR theory and it helps to predict the geometry of molecules.
According to this theory, electrons are arranged around the central atom of the molecule in such a way that there is minimum electrostatic repulsion between these electrons.
Now, calculate the total number of valence electrons in $SF_{6}$ .

$Valence\ electrons\ in\ SF_{6}= Valence\ electrons\ in\ S +\ 6(Valence\ electrons\ in\ F)$

Valence electrons of S = 6

Valence electrons of F = 7

Thus, the valence electrons in $SF_{6}$ are-

$Valence\ number\ of\ electrons\ in\ SF_{6} = (6) + 6(7) = 48\ electrons.$

The Lewis structure of $SF_{6}$ is - (Image attached).
In the structure, the number of atoms bonded to central atom (S) = 6.
Number of non-bonding electron pairs on the central atom = 0 (as all the valence electrons are bonded to F).
Electronic geometry in case of 6 bond pairs is octahedral.
Molecular geometry us also octahedral with bond angles 90°.
Central atom is sp3d2 hybridised.
$SF_{6}$ is a non-polar molecule.

To learn more about Lewis structures visit:

brainly.com/question/12307841?referrer=searchResults

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7 0

7 months ago

To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?

poizon [28]

We know that to relate solutions of with the factors of molarity and volume, we can use the equation: $M_{1} V_{1} = M_{2} V_{2}$

**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.

So now we can assign values to these variables. Let us say that the 18 M $H_{2} SO_{4}$ is the left side of the equation. Then we have:

$(18 M)(0.050 L)=(4.35M) V_{2}$

We can then solve for $V_{2}$ :

$V_{2}= \frac{(18M)(0.05L)}{4.35M}$ and $V_{2} =0.21 L$ or $210 mL$

We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.

7 0

3 years ago