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3 years ago

What is the mass of 1.25 L of ammonia gas at STP

Chemistry

1 answer:

Alex73 [517]3 years ago

7 0

Answer:

mass 1.25 Liters NH₃(gas) = 0.949 grams (3 sig-figs)

Explanation:

At STP (Standard Temperature-Pressure conditions => 0°C(=273K) and 1atm pressure, 1 mole any gas will occupy 22.4 Liters.

So, given 1.25 Liters ammonia gas at STP, convert to moles then multiply by formula wt. (17g/mole gives mass of NH₃.

moles NH₃(gas) = 1.25L NH₃(gas)/22.4L NH₄(gas)· NH₃(gas)mole⁻¹ = 0.0558 mole NH₃(gas).

Converting to grams NH₃(gas) = 0.0558 mole NH₃(gas) x 17 g·mol⁻¹ = 0.949 grams NH₃(gas).

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Olin [163]

This is an acid-base reaction where HF is the acid and H2O is the base (it's amphoteric and can be an acid or a base). The products would then H3O+ (the conjugate acid) and F- (the conjugate base). Now, we can simply construct a reaction using the found products and reactants. This acid-base reaction would be HF + H2O <--> H3O+ + F-.

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3 years ago

Rank the following 0.100 M solutions in order of increasing H3O+ concentration:

Inessa05 [86]

Answer:

HCN < HOCl < HF

Explanation:

The larger the Kₐ value, the stronger the acid.

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4 years ago

The mechanism for producing a concentrated urine involves all of the following except

avanturin [10]

Answer:

The correct answer is "obligatory water reabsorption in the proximal convoluted tubule".

Explanation:

The mechanism for producing concentrated urine cannot include the obligatory reabsorption of water in the proximal convoluted tubule since this process is part of the nephron, the system that filters the blood. Glucose and amino acids are reabsorbed almost entirely, as are approximately 70% of filtered potassium and 80% of bicarbonate.

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3 years ago

What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according

ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

Step 1: The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

Step 2: Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

Step 3: Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

Step 4: Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

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3 years ago