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2 years ago

How many molecules of NaCl are in 58.5 g NaCl?

Chemistry

1 answer:

Inga [223]2 years ago

7 0

Answer:

6.023*10 ²³ molecules

Explanation:

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

given mass of NaCl = w = 58.5 g

As we know the molecular mass of NaCl = m = 58.5 g/mol

Moles can be calculted from the above formula -

n = w / m = 58.5 g / 58.5 g/mol = 1 mol

Since , 1 mol has 6.023*10 ²³ particles ,

Therefore ,

1 mol of NaCl = 6.023*10 ²³ molecules

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If a piece of aluminum with a mass of 3.99 g and a temperature of 100.0 °C is dropped

Vinil7 [7]

Answer:

The final temperature of the system is 27.3°C.

Explanation:

Heat lost by aluminum = 3.99 × 0.91 × (100-T)

= 3.631 (100-T)

Heat gained by water = 10 × 4.184 × (T-21)

= 41.84 (T-21)

As,

Heat gained = Heat loss

or, 3.631(100-T) = 41.84(T-21)

or,363.1 - 3.631 T = 41.84 T - 878.64)

or, (41.84+ 3.631) T = 878.64 +363.1

or T= $\frac{1241.74}{45.47}$

or, T = 27.3°C

Hence the final temperature is 27.3°C.

8 0

3 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago

What is 25 m rounded to one significant figure

den301095 [7]

Rounded to 1 significant figure, 25 m would go to 30. This is because 0 isn't significant, so the 3 is the only significant figure.

8 0

3 years ago

Read 2 more answers

You observe 2 waves named alpha and beta. The alpha wave has a frequency of 5 Hz and the beta wave has a frequency of 2 Hz. Base

MAVERICK [17]

Answer:

shorter wavelength = alpha wave