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Aleksandr [31]
3 years ago
8

Calculate the molarity of a solution prepared by dissolving 6.80 grams of AgNO3 in 2.50 liters of solution.

Chemistry
1 answer:
tester [92]3 years ago
7 0

Answer:

0.016M

Explanation:

First we find the mole of AgNO3 by using the formula mass/molar mass..

Then we find molarity by the formula mole/volume...

I hope you get this..

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a proton is placed next to a negatively charged object in which direction does a proton move explain why
olganol [36]

The proton will move towards the object, since protons are positively charged.

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3 years ago
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Need help with 14 and 16 pls asap!! this is my friends test and im taking it tomorrow!!
marin [14]

Answer:

Q14: 17,140 g = 17.14 kg.

Q16: 504 J.

Explanation:

<u><em>Q14:</em></u>

  • To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).

m is the mass of the ice (m = ??? g).

c is the specific heat of the ice (c of ice = 2.1 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).

∵ Q = m.c.ΔT

∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)

∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.

<u><em>Q16:</em></u>

  • To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by ice (Q = ??? J).

m is the mass of the ice (m = 12.0 g).

c is the specific heat of the ice (c of ice = 2.1 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).

∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.

6 0
3 years ago
3. Matter with a composition that is always the same is a(n)<br> substance<br> mixture<br> compound
Andre45 [30]

Answer:

substance

Explanation:

A pure substance is a form of matter that has a constant composition (meaning it's the same everywhere) and properties that are constant throughout the sample (meaning there is only one set of properties such as melting point, color, boiling point, etc

7 0
3 years ago
1. a.) Calculate the wavelength of light which has a frequency of 5.25 x 10 14 Hz.
Scilla [17]
<h3>Answer:</h3>

5.71 × 10² nm

<h3>Explanation:</h3>

The product of wavelength and frequency of a wave gives the speed of the wave.

Therefore;

Velocity of wave = Wavelength × Frequency

c = f ×λ

In our case;

Frequency = 5.25 × 10^14 Hz

Speed of light = 2.998 × 10^8m/s

But;

λ = c ÷ f

  = 2.998 × 10^8m/s ÷  5.25 × 10^14 Hz

  = 5.71 × 10^-7 m

But; 1 M = 10^9 nm

Therefore;

wavelength = 5.71 × 10^-7 × 10^9

                  = 5.71 × 10² nm

The wavelength of light wave 5.71 × 10² nm

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P=18000000/6 zeros. not sure how to do rest

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