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2 years ago

2.00 g of an unknown gas at STP fills a 500. mL flask. What is the molar mass of the gas?

Chemistry

1 answer:

otez555 [7]2 years ago

4 0

Answer:

100g/mol

Explanation:

Given parameters:

Mass of unknown gas = 2g

Volume of gas in flask = 500mL = 0.5dm³

Unknown:

Molar mass of gas = ?

Solution:

Since we know the gas is at STP;

1 mole of substance occupies 22.4dm³ of space at STP

Therefore,

0.5dm³ will have 0.02mole at STP

Now;

Number of moles = $\frac{mass}{molar mass}$

Molar mass = $\frac{mass}{number of moles}$ = $\frac{2}{0.02}$ = 100g/mol

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Choose the answer in which the three atoms and/or ions are listed in order of increasing EXPECTED size (smallest particle listed

svp [43]

Answer:

Ar < Cl - < S2-

Explanation:

All the species written above are isoelectronic. This means that they all possess the same number of electrons. All the species above possess 18 electrons, the noble gas electron configuration.

However, for isoelectronic species, the greater the atomic number of the specie, the smaller it is. This is because, greater atomic number implies that their are more protons in the nucleus exerting a greater attractive force on the electrons thereby making the specie smaller in size due to high electrostatic attraction.

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2 years ago

If 10.0 grams of NaHCO3 is added to 10.0 g of HCl, determine the efficiency of baking soda as an antacid if 6.73 g of NaCl was p

Lapatulllka [165]

Answer:

percentage yield of NaCl = 96.64%

Explanation:

The reaction was between NaHCO3 and HCl .The chemical equation can be represented below:

NaHCO3 + HCl → NaCl + H2O + CO2 . The balance equation is

NaHCO3 + HCl → NaCl + H2O + CO2

The question ask us to calculate the percentage yield of NaCl.

The efficiency of NaHCO3 as an antacid , the limiting reactant is NaHCO3

1 mole of NaHCO3 produces 1 mole of NaCl

Therefore,

molar mass of NaHCO3 = 23 +1 + 12 + 48 = 84 g

molar mass of NaCl = 23 + 35.5 = 58.5 g

1 mole of NaHCO3 = 84 g

1 mole of NaCl = 58.5 g

since 84 g of NaHCO3 produces 58.5 g of NaCl

10 g of NaHCO3 will produce ? grams of NaCl

cross multiply

Theoretical yield of NaCl = (10 × 58.5)/84

Theoretical yield of NaCl = 585/84

Theoretical yield of NaCl = 6.9642857143 g

percentage yield of NaCl = actual yield/theoretical yield × 100

percentage yield of NaCl = 6.73/6.9642857143 × 100

percentage yield of NaCl = 673/6.9642857143

percentage yield of NaCl = 96.635897436%

percentage yield of NaCl = 96.64%

3 0

2 years ago

Read 2 more answers

In the equilibrium system described by: PO43-(aq) + H2O(1) = HPO42-(aq) + OH-(aq) Brønsted-Lowry theory would designate: A) PO43

Kamila [148]

Answer:

The correct option is: B) H₂0 and OH⁻ as a conjugate pair

Explanation:

According to Brønsted-Lowry theory, the acids are the chemical substances that form a conjugate base by donating a proton and bases are the chemical substances that form conjugate acid by accepting a proton.

In the given chemical reaction: PO₄³⁻(aq) + H₂O(l) ⇄ HPO₄²⁻(aq) + OH⁻(aq)

According to Brønsted-Lowry theory, PO₄³⁻ and OH⁻ are bases. Whereas, H₂O and HPO₄²⁻ are acids.

Also, PO₄³⁻ and HPO₄²⁻ are the conjugate acid-base pair; and H₂O and OH⁻ are the conjugate acid-base pair.

6 0

2 years ago

Dehydrohalogenation of tert-pentyl bromide at higher temperatures will produce 2-methyl-1-butene as the chief product when

salantis [7]

I'm going to suppose you want the adjusted chemical reaction, using the formulas of the compounds. You can see it in the image attached.

6 0

2 years ago

Anyway, which is a spectator ion involved in the reaction of k2cro4(aq) and ba(no3)2(aq)?

marissa [1.9K]

Nitrate ions $\text{NO}_3^{-}$ and potassium ions $\text{K}^{+}$ .

<h3>Explanation</h3>

Potassium dichromate undergoes a double-decomposition reaction with barium nitrate to produce barium dichromate and potassium nitrate. The reaction is possible due to the low solubility of barium dichromate that precipitate out of the solution readily after its production.

Start with the balanced chemical equation for this process:

$\text{K}_2\text{CrO}_4 \; (aq) + \text{Ba}(\text{NO}_3)_2 \; (aq) \to \text{Ba}(\text{CrO}_4)_2 \; (s) + 2 \; \text{KNO}_3 \; (aq)$

Rewrite the chemical equation as an ionic one; express all soluble salts- those with state symbol (aq)- as their constituting ions while leaving the insoluble (s) intact.

$2 \; \text{K}^{+}\; (aq)+ 2\; \text{CrO}_4^{-} \; (aq) + \text{Ba}^{2+} \; (aq) + 2 \; \text{NO}_3^{-} \; (aq) \\ \to \text{Ba}(\text{CrO}_4)_2 \; (s) + 2 \; \text{K}^{+} \; (aq) + 2 \; \text{NO}_3^{-} \; (aq)$

$\text{K}^{+}$ and $\text{CrO}_4^{-}$ are found on both sides of the equation by the same quantity. The two ions thus took no part in the net reaction and act as spectator ions.

5 0

2 years ago