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3 years ago

Why didn't scientists know about atoms for a long time

1 answer:

6 0

They didnt have a powerfull enough microscope to see them an when they could they were moving to fast unless they were cold

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a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill

baherus [9]

Answer:

1223.38 mmHg

Explanation:

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

Also,

Moles = mass (m) / Molar mass (M)

Density (d) = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

$PM=dRT$

Given that:-

d = 1.80 g/L

Temperature = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (32 + 273.15) K = 305.15 K

Molar mass of nitrogen gas = 28 g/mol

Applying the equation as:

P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

⇒P = 1223.38 mmHg

<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>

5 0

3 years ago

The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?

sweet [91]

Answer : The maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

Explanation :

The solubility equilibrium reaction will be:

$Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-$

Initial conc. 0.220 0

At eqm. (0.220+s) 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ni^{2+}][CN^-]^2$

Now put all the given values in this expression, we get:

$3.0\times 10^{-23}=(0.220+s)\times (2s)^2$

$s=5.84\times 10^{-12}M$

Therefore, the maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

7 0

3 years ago

What is the osmotic pressure in atm of

Dmitrij [34]

Answer:

12.2329 atm

Explanation:

Applying

PV = nRT.................... Equation 1

Make P the subject of the equation

P = (n/V)RT................ Equation 2

Where P = Osmotic pressure, (n/V) = Molarity of urea, T = Temperature, R = molar gas constant.

From the question,

Given: (n/V) = 0.5 M, T = 25°C = (273+25) = 298 K, R = 0.0821 Latm/mol.K

Substitute these values into equation 2

P = (0.5)(298)(0.0821)

P = 12.2329 atm

5 0

3 years ago

A chemistry student needs 45.0mL of pentane for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the stud

Masja [62]

The mass of pentane the student should weigh out is

The density of pentane is 0.626 gcm-3

To calculate the mass of pentane following expression is used,

(Density is defined as the mass divide by volume)

Density = mass / volume

mass of pentane = Density of pentane * Volume of pentane

mass of pentane = 0.626 gcm-3 * 45.0 mL

= 28.17 g

Here the unit of mass of pentane is g,

However the unit of density is gcm-3 and unit of volume is mL i.e. cm3

Hence, Mass = gcm-3 * cm3

Mass = g

The mass of pentane the student should weigh out is 28.17g

Learn more about Density on

brainly.com/question/1354972

#SPJ1

5 0

2 years ago

2. What is the mass of one mole of Na?

borishaifa [10]

Sodium, Atomic mass: 22.989769 g
You can see in a periodic table

5 0

3 years ago