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Oliga [24]
3 years ago
15

A solution is prepared by dissolving 15.0 g of nh3 in 250.0 g of water. the density of the resulting solution is 0.974 g/ml. the

molality of nh3 in the solution is
Chemistry
1 answer:
sergeinik [125]3 years ago
6 0
The formula for molality---> m = moles solute/ Kg of solvent

the solute here is NH₃ because it's the one with less amount. which makes water the solvent.

1) let's convert the grams of NH₃ to moles using the molar mass

molar mass of NH₃= 14.0 + (3 x 1.01)= 17.03 g/ mol

15.0 g (1 mol/ 17.03 g)= 0.881 mol NH₃

2) let's convert the grams of water into kilograms (just divide by 1000)

250.0 g= 0.2500 kg

3) let's plug in the values into the molality formula

molality= mol/ Kg---> 0.881 mol/ 0.2500 kg= 3.52 m
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why do you use a graduated cylinder to measure out the desired volume of koh and h2so4, rather than a pipet or a buret?
sergey [27]

The graduated cylinder is used to measure the volume of KOH and H2SO4 when accurate volume measurement is not required.

In the laboratory certain graduated apparatus are used to measure liquids. These graduated apparatus used to measure liquids include;

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Sometimes, we are not really looking for a strictly accurate volume of liquid and we can use a graduated cylinder to measure the volume of liquid in such cases.

However, when we need to have strictly accurate volume measurement, we need a pipet or a buret.

Learn more: brainly.com/question/15670537

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2 years ago
Which ion is formed when an atom of fluorine (f) gains one electron? f–1 f–2 f 1 f 2
gregori [183]

Answer:

If a fluorine atom gains an electron, it becomes a fluoride ion with an electric charge of -1.

Explanation:

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6 0
2 years ago
How many moles are present in a 24.5 gram sample of K2Cr2O7?
attashe74 [19]
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6 0
3 years ago
6 Bumper cars are a favorite ride at amusement parks. When bumper cars hit, a force is exerted on both cars, and they move in op
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2 years ago
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The half-life of Co-59 is 35 days. How much of a 610.0-gram sample will remain after 140 days?
vitfil [10]

Answer:

Amount left after 140 days is 38.125 g.

Explanation:

Given data:

Half life of Co-59 = 35 days

Total mass of sample = 610.0 g

Sample remain after 140 days = ?

Solution:

Number of half lives passes during 140 days:

Number of half lives  = T elapsed / Half lives

Number of half lives  = 140 days / 35 days

Number of half lives  = 4

Amount left:

At time zero = 610 g

At first half life = 610 g/2 = 305 g

At second half life = 305 g/2 = 152.5 g

At 3rd half life = 152.5 g/2 = 76.25 g

At 4th half life = 76.25 g /2 =38.125 g

Amount left after 140 days is 38.125 g.

3 0
3 years ago
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