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2 years ago

How many atoms are in 3.2 pg of Ca? The molar mass of Ca is 40.08 g/mol.

Chemistry

1 answer:

anyanavicka [17]2 years ago

7 0

Answer:

4.81×10¹⁰ atoms.

Explanation:

We'll begin by converting 3.2 pg to Ca to grams (g). This can be obtained as follow:

1 pg = 1×10¯¹² g

Therefore,

3.2 pg = 3.2 pg × 1×10¯¹² g / 1 pg

3.2 pg = 3.2×10¯¹² g

Therefore, 3.2 pg is equivalent to 3.2×10¯¹² g

Next, we shall determine the number of mole in 3.2×10¯¹² g of Ca. This can be obtained as follow:

Mass of Ca = 3.2×10¯¹² g

Molar mass of Ca = 40.08 g/mol

Mole of ca=.?

Mole = mass /molar mass

Mole of Ca = 3.2×10¯¹² / 40.08

Mole of Ca = 7.98×10¯¹⁴ mole.

Finally, we shall determine the number of atoms present in 7.98×10¯¹⁴ mole of Ca. This can be obtained as illustrated below:

From Avogadro's hypothesis,

1 mole of Ca contains 6.02×10²³ atoms.

Therefore, 7.98×10¯¹⁴ mole of Ca will contain = 7.98×10¯¹⁴ × 6.02×10²³ = 4.81×10¹⁰ atoms.

Therefore, 3.2 pg of Ca contains 4.81×10¹⁰ atoms.

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How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 163 mL of coffee at 97.9°C so that the r

Stolb23 [73]

Answer:

248 mL

Explanation:

According to the law of conservation of energy, the sum of the heat absorbed by water (Qw) and the heat released by the coffee (Qc) is zero.

Qw + Qc = 0

Qw = -Qc [1]

We can calculate each heat using the following expression.

Q = c × m × ΔT

where,

c: specific heat

m: mass

ΔT: change in the temperature

163 mL of coffee with a density of 0.997 g/mL have a mass of:

163 mL × 0.997 g/mL = 163 g

From [1]

Qw = -Qc

cw × mw × ΔTw = -cc × mc × ΔTc

mw × ΔTw = -mc × ΔTc

mw × (54.0°C-25.0°C) = -163 g × (54.0°C-97.9°C)

mw × 29.0°C = 163 g × 43.9°C

mw = 247 g

The volume corresponding to 247 g of water is:

247 g × (1 mL/0.997 g) = 248 mL

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3 years ago

KOH (aq) + HNO(little3) (aq) = KNO(little 3) (aq) + H20 (1)

Yuki888 [10]

The reaction already balanced

<h3>Further explanation </h3>

Equalization of chemical reactions can be done using variables. Steps in equalizing the reaction equation:

• 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c, etc.

• 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index (subscript) between reactant and product

• 3. Select the coefficient of the substance with the most complex chemical formula equal to 1

For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:

Balancing C, H and the last O

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KOH (aq) + HNO₃ (aq) ⇒ KNO₃ (aq) + H₂0 (l)

give coefficient

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K : left = a, right = 1⇒a=1

N : left = b, right =1, b=1

H: left = a, right a+b=2c⇒1+1=2c⇒2=2c⇒c=1

KOH (aq) + HNO₃ (aq) ⇒ KNO₃ (aq) + H₂0 (l)⇒ already balanced

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3 years ago

How can you distinguish between crystalline allotropic modifications of Sulphur from those of amorphous allotrops?

Burka [1]

The crystalline allotropes of sulfur are very strong and have a high melting and boiling point while the amorphous allotropes of sulfur are brittle and breaks easily.

<h3>What is a crystalline substance?</h3>

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Also, the crystalline allotropes of sulfur are very strong and have a high melting and boiling point while the amorphous allotropes of sulfur are brittle and breaks easily.

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1 year ago

Which properties cause chemical reactions?

GREYUIT [131]

D. Chemical and Physical Properties

3 0

3 years ago

Read 2 more answers

1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.

Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

$\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}$

$\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}$

$\Delta \ S = {(1*37.7)In \dfrac{263}{273}$

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

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2 years ago