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3 years ago

A 4,000 mL solution of AgNO3 contains 17.00 g of solute in water. Calculate the molar concentration of the solution.

Chemistry

2 answers:

MariettaO [177]3 years ago

6 0

The answer is 0.03 M

blsea [12.9K]3 years ago

3 0

The molarity is moles/liters.

First, convert 4,000 mL to L:

4000 mL --> 4 L

Now, you must convert the 17 g of solute to moles by dividing the number of grams by the molar mass. The molar mass of AgNO3 is 169.87 g/mol:

17 / 169.87 = .1

Now that you have both the number of moles and the liters, plug them into the initial equation of moles/liters:

.1/4 = .025

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Answer:

The correct answer is electrons.

Explanation:

The electrons refer to the negatively charged constituents, which are arranged in the orbits around the nucleus of an atom. It determines all the chemical and physical characteristics of an atom except the radioactivity and mass. It is a stable subatomic particle witnessed in all the atoms and functioning as the chief carrier of electricity in solids.

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A 40.2 g sample of a metal heated to 99.3Â°C is placed into a calorimeter containing 120 g of water at 21.8Â°C. The final temper

aliina [53]

Answer:

B) Iron (c=0.45 J/g°C)

Explanation:

Given that:-

Heat gain by water = Heat lost by metal

Thus,

$m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

$m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)$

For water:

Mass = 120 g

Initial temperature = 21.8 °C

Final temperature = 24.5 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 40.2 g

Initial temperature = 99.3 °C

Final temperature = 24.5 °C

Specific heat of metal = ?

So,

$120\times 4.184\times (24.5-21.8)=40.2\times C_{metal}\times (99.3-24.5)$

$40.2C_{metal}\left(99.3-24.5\right)=120\times \:2.7\times \:4.184$

$40.2C_{metal}\left(99.3-24.5\right)=1355.616$

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This value corresponds to iron. Thus answer is B.

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3 years ago

Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer

bogdanovich [222]

Answer:

Explanation:

This can be contradictory, depending on whether the 0.1 M

is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

VA− = 9.125 mL

VHA = 15.875 mL

The Henderson-Hasselbalch equation is:

pH = pKa + log [A−][HA]

We have a pH 4.5

solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

[A−][HA]=10

pH − pKa = 104.5 − 4.74 = 0.5754

Now, if the total concentration is

0.10 M , then:

[HA] + [A−] 0.5754

[HA] = 0.10 M

⇒[HA] = 0.10 M 1.0000 +0.5754

= 0.0635 M

−−−−−−−−

⇒[A−] = 0.0365 M

−−−−−−−−

and these concentrations are AFTER mixing. Since the total volume is 50 mL , or 0.050 L, the mols of each component (which are constant!) are:

nA − = 0.0365 molL × 0.050L =

0.001825 mols

−−−−−−−−−−−−

nHA = 0.0635 molL × 0.050L =

0.003175 mols

−−−−−−−−−−−−

So, if both of the starting concentrations were

0.20 M, we can find the volume they each start with:

VA − = 1 L0.20mols

A− × 0.001825mols A− = 0.009125 L = 9.125 mL

−−−−−−−−

VHA = 1 L 0.20 mols HA × 0.003175

mols HA = 0.015875 L = 15.875 mL

−−−−−−−−−

And this should make sense, because the total starting volume is

25.000 mL , the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.

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2 years ago

the reversible reaction N2(g) 3H2(g) <--> 2NH3(g) produces ammonia, which is a fertilizer. At equilibrium, a 1L flask cont

Alik [6]

Answer:

Explanation:

the chemical equilibrium constant can be easily calculated since the concentrations at equilibrium are given.the calculation shows the value of Kc for the reversible reaction and forward reaction

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3 years ago