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3 years ago

26.8 moles of argon contain how many atoms?

Chemistry

2 answers:

Maru [420]3 years ago

4 0

<span>1.61 × 1023 Multiply by 26.8 to get the answer.161.33 x 10 ^23 </span>

Alisiya [41]3 years ago

3 0

Answer : The number of atoms of argon are $1.61\times 10^{25}$

Explanation : Given,

Moles of argon = 26.8 moles

As we know that,

1 mole of substance always contains $6.022\times 10^{23}$ number of atoms.

As, 1 mole of argon contains $6.022\times 10^{23}$ number of atoms

So, 26.8 mole of argon contains $26.8\times 6.022\times 10^{23}=1.61\times 10^{25}$ number of atoms.

Therefore, the number of atoms of argon are $1.61\times 10^{25}$

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A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad

ASHA 777 [7]

$0 \; \textdegree{\text{C}}$

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

$E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ}$ and
$m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}$

Let the final temperature of the system be $t \; \textdegree{\text{C}}$ . Thus $\Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial}) - t_{0} = t \; \textdegree{\text{C}}$

$Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ}$ (converted to kilojoules)
$Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}$
$Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}$

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable $t$ .

$E(\text{absorbed} ) = E(\text{released})$

$E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})$
$E(\text{released}) = Q(\text{lead})$

Confirm the uniformity of units, equate the two expressions and solve for $t$ :

$66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)$

$t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}}$ which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., $t \le 0 \; \textdegree{\text{C}}$ . However, there's no way for the temperature of the system to go below $0 \; \textdegree{\text{C}}$ ; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at $0 \; \textdegree{\text{C}}$ ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

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3 years ago

The pressure of a fixed mass of gas is increased from 100 kPa to 600 kPa at a constant temperature. The new volume of the gas is

Dima020 [189]

Answer: The new volume of the gas is smaller.

Explanation:

Volume and pressure are inversely proportional (as one goes up, one goes down). So as you increase the pressure, you decrease the volume.

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Calculate the amount of heat needed to boil 159.g of water

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Okay I did the math and I'm guessing around 18*C

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Match the compounds as organic (O) or inorganic (I).

SashulF [63]

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3 years ago

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When a mercury-202 nucleus is bombarded with a neutron, a proton is ejected. What element is formed?

Galina-37 [17]

That will make a gold-202 nucleus.

<h3>Explanation</h3>

Refer to a periodic table. The atomic number of mercury Hg is 80.

Step One: Bombard the $\displaystyle ^{202}_{\phantom{2}80}\text{Hg}$ with a neutron $^{1}_{0}n$ . The neutron will add 1 to the mass number 202 of $^{202}_{\phantom{2}80}\text{Hg}$ . However, the atomic number will stay the same.

New mass number: 202 + 1 = 203.
Atomic number is still 80.

$^{202}_{\phantom{2}80}\text{Hg} + ^{1}_{0}n \to ^{203}_{\phantom{2}80}\text{Hg}$ .

Double check the equation:

Sum of mass number on the left-hand side = 202 + 1 = 203 = Sum of mass number on the right-hand side.
Sum of atomic number on the left-hand side = 80 = Sum of atomic number on the right-hand side.

Step Two: The $^{203}_{\phantom{2}80}\text{Hg}$ nucleus loses a proton $^{1}_{1}p$ . Both the mass number 203 and the atomic number will decrease by 1.

New mass number: 203 - 1 = 202.
New atomic number: 80 - 1 = 79.

Refer to a periodic table. What's the element with atomic number 79? Gold Au.

$^{203}_{\phantom{2}80}\text{Hg} \to ^{202}_{\phantom{2}79}\text{Au} + ^{1}_{1}p$ .

Double check the equation:

Sum of mass number on the left-hand side = 203 = 202 + 1 = Sum of mass number on the right-hand side.
Sum of atomic number on the left-hand side = 80 = 79 + 1 = Sum of atomic number on the right-hand side.

A gold-202 nucleus is formed.

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3 years ago