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3 years ago

Wavelengths are often represented in nm. What is the diameter of a helium (He) atom in

Chemistry

2 answers:

Wittaler [7]3 years ago

8 0

Answer: The diameter of helium atom is 0.1 nm.

Explanation:

We are given:

Diameter of helium atom = $1.0\times 10^{-13}km$

To convert this diameter from kilometers to nano meters, we use the conversion factor:

$1km=10^{12}nm$

Converting the above diameter into nano meters, we get:

$\Rightarrow (1.0\times 10^{-13}km)\times (\frac{10^{12}nm}{1km})\\\\\Rightarrow 0.1nm$

Hence, the diameter of helium atom is 0.1 nm

lianna [129]3 years ago

5 0

Answer:

Diameter He = 0,1 nm.

Explanation:

Km to nm:

⇒ Diameter He = 1.0 E-13 Km * ( 1000 m / Km ) * ( 1 E9 nm / m )

⇒ Diameter He = 0.1 nm

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Why do oxygen and sulfur have similar chemical properties?

Step2247 [10]

Answer:

Oxygen and sulfur have similar chemical properties because both elements have six electrons in their outermost electron shells. Indeed, both oxygen and sulfur form molecules with two hydrogen atoms: water and hydrogen sulfide

Explanation:

3 0

3 years ago

A student dissolves 15.0 g of ammonium chloride(NH4Cl) in 250. 0 g of water in a well-insulated open cup. She then observes the

iren2701 [21]

Answer:

1) Endothermic.

2) $Q_{rxn}=4435.04J$

3) $\Delta _rH=15.8kJ/mol$

Explanation:

Hello there!

1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.

2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

$Q_{rxn}=-(15.0g+250.0g)*4.184\frac{J}{g\°C}(16.0-20.0)\°C\\\\ Q_{rxn}=4435.04J$

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:

$\Delta _rH=\frac{ Q_{rxn}}{n}\\\\\Delta _rH= \frac{ 4435.04J}{15.0g*\frac{1mol}{53.49g} } *\frac{1kJ}{1000J} \\\\\Delta _rH=15.8kJ/mol$

Best regards!

4 0

3 years ago

A flask with a volume of 125.0 mL contains air with a density of 1.298 g/L. what is the mass of the air contained in the flask?

horsena [70]

A flask with a volume of 125.0 mL contains air with a density of 1.298 g/L. what is the mass of the air contained in the flaskThe given are: 
1. Mass = ?
2. Density = 1298 g/L
3. Volume = 125mL to L
a. 125 ml x 0.001l/1ml = 0.125 L

Formula and derivation: 
1. density = mass / volume 
2. mass = density / volume

Solution for the problem: 

1. mass = 1298 g/L / 0.125 L = 10384g

8 0

3 years ago

Ammonia, nh3, for fertilizer is made by causing hydrogen and nitrogen to react at high temperature and pressure. How many moles

charle [14.2K]

Answer:

$0.30molNH_3$

Explanation:

Hello there!

In this case, since the reaction for the formation of ammonia is:

$3H_2+N_2\rightarrow 2NH_3$

We can evidence the 1:2 mole ratio of nitrogen gas to ammonia; therefore, the appropriate stoichiometric setup for the calculation of the moles of the latter turns out to be:

$0.15molN_2*\frac{2molNH_3}{1molN_2}$

And the result is:

$0.30molNH_3$

Best regards!

7 0

3 years ago

Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a

coldgirl [10]

The final temperature, t₂ = 30.9 °C

<h3>Further explanation</h3>

Given

24.0 kJ of heat = 24,000 J

Mass of calorimeter = 1.3 kg = 1300 g

Cs = 3.41 J/g°C

t₁= 25.5 °C

Required

The final temperature, t₂

Solution

Q = m.Cs.Δt

Q out (combustion of compound) = Q in (calorimeter)

24,000 = 1300 x 3.41 x (t₂-25.5)

t₂ = 30.9 °C

3 0

3 years ago

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