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3 years ago

The mole is most similar to:

2 answers:

8 0

The best and most correct answer among the choices provided by your question is the third choice or letter C.

<span>The mole is most similar to a dozen which is a measure of number.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

nekit [7.7K]3 years ago

4 0

Answer : The correct option is, a dozen (a measure of number).

Explanation :

Mole : It is defined as the amount of substance that contains Avogadro number of the substance.

Avogadro's number is given by $6.022\times 10^{23}$ . That means,

$1\text{ mole}=6.022\times 10^{23}$

Dozen : It is defined as the amount of substance that contains 12 number of particles.

$1\text{ dozen}=12\text{ objects}$

Hence, the mole is most similar to a dozen (a measure of number).

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In body cells such as skin and nerve cells, chromosomes are found in _____. A. singlets B. pairs C. triplets D. quadruplets plea

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The answer is B. Pairs

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4 years ago

Read 2 more answers

A truckload of apples weighs 2550 lbs. If 5 (exact) apples weigh 1250 g, how many apples are in the truck?

egoroff_w [7]

Answer:

There are 4,626 apples in the truck.

Explanation:

Mass of 5 apples = 1250

Mass of an apple = $\frac{1250 g}{5}=250 g$

Mass of apples in the truck = 2550 lbs

1 lbs = 453.592 g

$2550 lbs=2550\times 453.592 g$

Let the number of apples with mass of 2550 lbs be a.

$a\times 250 g=2550\times 453.592 g$

$a=\frac{2550\times 453.592 g}{250 g}=4,626.63$

4,626.63 ≈ 4,626

So, there are 4,626 apples in the truck.

3 0

3 years ago

What is the completed balanced reaction for the replacement Al×H2SO4

dsp73

Answer:

2 Al(s) + 3 H₂SO₄(aq) ⇒ Al₂(SO₄)₃(aq) + 3 H₂(g)

Explanation:

Let's consider the unbalanced equation for the single replacement reaction between Al and H₂SO₄.

Al(s) + H₂SO₄(aq) ⇒ Al₂(SO₄)₃(aq) + H₂(g)

We will begin balancing S atoms by multiplying H₂SO₄ by 3.

Al(s) + 3 H₂SO₄(aq) ⇒ Al₂(SO₄)₃(aq) + H₂(g)

Then, we balance Al atoms by multiplying Al by 2.

2 Al(s) + 3 H₂SO₄(aq) ⇒ Al₂(SO₄)₃(aq) + H₂(g)

Finally, we get the balanced equation by multiplying H₂ by 3.

2 Al(s) + 3 H₂SO₄(aq) ⇒ Al₂(SO₄)₃(aq) + 3 H₂(g)

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3 years ago

Consider the reaction A ⟶Products

shtirl [24]

The reaction is zero order in [A].

Doubling [A] from 0.10 mol·L⁻¹ to 0.20 mol·L⁻¹ has no effect on the initial rate.

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4 years ago

The concentration of chlorobenzene (C&HsCl) in water is 100 mol/m3. density is 1.00 g/cm3 The solution (a) What is the weigh

muminat

Answer:

Part 1

(a) 0.0113

(b) 11300 ppm

(d) 0.100 M

(e) 0.101 m

Part 2

(a) 9.45 *10⁻³

(b) mole fraction = 2.45 *10⁻³

Explanation:

Chlorobenzene formula is C₆H₅Cl

Part 1: We are given a concentration of chlorobenzene in water of 100 mol/m³, and a density of the solution of 1.00 g/cm³.

(a) weight fraction C₆H₅Cl = mass C₆H₅Cl / mass solution

We know there are 100 moles of C₆H₅Cl per m³ of solution.

To get the mass of C₆H₅Cl we'll convert the moles to mass by using the molar mass:

Molar mass C₆H₅Cl = 6*12.011 + 5*1.00794 + 35.4527 = 112.558 g/mol

mass C₆H₅Cl = moles C₆H₅Cl * molar mass C₆H₅Cl

mass C₆H₅Cl = 100 moles * 112.558 g/mol = 11255.8 g

11255.8 g of C₆H₅Cl are in 1 m³ of solution.

Next we'll convert 1 m³ of solution to mass by using the density

mass solution = volume solution * density of solution

$mass solution = 1m^{3} *\frac{(100cm)^{3} }{ 1m^{3}} * \frac{1.00 g}{cm^{3} } = 1.00 *10^{6} g$

weight fraction C₆H₅Cl = 11256 g / 1.00 *10⁶ g = 0.0113

(b) ppm stands for "parts per million" and it is usually expressed as mg per Liter of solution

We already calculated that there are 11256 g or more exactly 11300 g of C₆H₅Cl in 1 m³ of solution, so lets convert to mg/L:

$\frac{11300 g}{1 m^{3} } * \frac{1000 mg}{1 g} * \frac{1 m^{3} }{1000 L} = 11300 mg/L$

So the solution is 11300 ppm

(c) mole fraction = moles of C₆H₅Cl / total moles in solution

total moles = moles C₆H₅Cl + moles water

moles water = mass water / molar mass water

mass water = mass solution - mass C₆H₅Cl

moles of C₆H₅Cl = 100 moles

mass water = 1.00 *10⁶ g of solution - 11256 g = 988744 g of water

moles water = 988744 g / 18.0153 g/mol = 54884 moles water

total moles = 100 + 54884 = 54984 moles

mole fraction = 100 moles of C₆H₅Cl / 54984 moles = 1.82 *10⁻³

(d) Molarity = moles C₆H₅Cl / Liters of solution

We know the solution is 100 mol / m³ so we just have to convert the m³ to L:

$\frac{100 mol}{m^{3} } * \frac{1 m^{3}}{1000 L} = 0.100 mol / L = 0.100 M$

(e) Molality = moles C₆H₅Cl / kg water

We know that there are 100 moles per 988744 g of water, so we need to convert the grams of water to kilograms.

$Molality = \frac{100 moles}{988744 g} *\frac{1000 g}{1 kg} = 0.101 m$

____________________________________

Part 2: Concentration of C₆H₅Cl in air is 0.100 mol/m³, at 25 °C and 1 atm.

Molar mass air = 28.84 g/mol

(a) weight fraction C₆H₅Cl = mass C₆H₅Cl / total mass

mass C₆H₅Cl = 0.100 mol * 112.558 g/mol = 11.26 g

total mass = mass C₆H₅Cl + mass air

mass air = moles air * molar mass air

moles air = total moles - moles C₆H₅Cl

We can calculate the total moles by using the ideal gas law:

P V = n R T

where P is pressure in atm, V is volume in L, n is the number of moles, R is the gas constant and T is temperature in Kelvin.

n = P V / R T

P = 1 atm

V = 1 m³ = 1000 L

R = 0.08206 L atm K⁻¹ mol⁻¹

T = 25 + 273.15 = 298 K

n = (1 atm * 1000 L) / (0.08206 L atm K⁻¹ mol⁻¹ * 298 K) = 40.89 moles

moles air = 40.89 - 0.100 = 40.79 moles air

mass air = 40.79 mol * 28.84 g/mol = 1176.4 g

total mass = 1176.4 g + 11.26 g = 1188 g

weight fraction = 11.26 g / 1188 g = 9.45 *10⁻³

(b) mole fraction = moles C₆H₅Cl / total moles

mole fraction = 0.100 / 40.89 = 2.45 *10⁻³

(c) ppm = mg C₆H₅Cl / Liters

We already know there are 11.26 g C₆H₅Cl in 1 m³, which is the same as 1000 L, so:

$\frac{11.26 g}{1000 L} *\frac{1000 mg}{1 g} = 11.3 mg/L$

The concentration is 11.3 ppm

8 0

3 years ago