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Alenkinab [10]
3 years ago
14

It takes 2,500,00 Liters of Helium to fill the Goodyear Blimp. How many moles is this?

Chemistry
1 answer:
Valentin [98]3 years ago
3 0

Answer:

102.26 moles of helium were required to Fill the Goodyear Blimp

Explanation:

To solve this question we need to use combined gas law:

PV = nRT

<em>Where P is pressure, V is volume of gas (2500L), n are moles of gas (Our incognite), R is gas constant (0.082atmL/molK) and T is absolute temperature</em>

<em />

Assuming atmospheric condition we can write P = 1atm and T = 25°C = 298.15K

Replacing:

PV/RT = n

1atm*2500L / 0.082atmL/molK*298.15K = n

<h3>102.26 moles of helium were required to Fill the Goodyear Blimp</h3>

<em />

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An alpha particle is equivalent to the nucleus of an atom of which element?(1 point)
Katyanochek1 [597]

Answer:

this is what i got

Explanation:

α-decay: When a radioactive nucleus disintegrates by emitting an αα-particle, the atomic number decreases by two and mass number decreases by four. Example: 88​Ra226→86​Rn222+2​He4.

3 0
3 years ago
4th one<br> Mark the brainliest
ale4655 [162]

Answer:

1. The electronic configuration of X is: 1s2 2s2 sp6 3s2

2. The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6

3. The formula of the compound form by X and Y is given as: XY

Explanation:

For X to loss two electrons, it means X is a group 2 element. X can be any element in group 2. The electronic configuration of X is:

1s2 2s2 sp6 3s2

To get the electronic configuration of the anion of element Y, let us find the configuration of element Y. This is done as follows:

Y receives two electrons from X to complete its octet. Therefore Y is a group 6 element. The electronic configuration of Y is given below

1s2 2s2 2p4

The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6

The formula of the compound form by X and Y is given below :

X^2+ + Y^2- —> XY

Their valency will cancel out thus forming XY

4 0
4 years ago
A florist prepares a solution of nitrogen-phosphorus fertilizer by dissolving 5.66 g of NH₄NO₃ and 4.42 g of (NH₄)₃ PO₄ in enoug
Alexandra [31]

In finding the molarity of a solution, we use the following formula:

M=moles soluteL solution

What is Molarity?

The number of moles of the solute is calculated by dividing the mass of the solute by its molar mass.

<h3 />

The molar mass of  NH4NO3 and (NH4)3PO4 are  80.043 g/mol and 149.0867 g/mol, respectively.

molesNH+4inNH4NO3=5.66 g80.043 g/mol×1molNH+41molNH4NO3=0.0707 mol

molesNH+4in(NH4)3PO4=4.42 g149.0867 g/mol×3molNH+41mol(NH4)3PO4=0.0889 mol

total molesNH+4=0.0707 mol+0.0889 mol=0.1596 mol

molesPO3−4in(NH4)3PO4=4.42 g149.0867 g/mol×1molPO3−41mol(NH4)3PO4=0.0296 mol

[NH+4]=0.1596 mol20.0 L=7.98×10−3 M NH+4

[PO3−4]=0.0296 mol20.0 L=1.48×10−3 M PO3−4

Therefore, [PO3−4]\\ has a molarity of  1.48×10−3 M PO3−

To learn more about Molarity click on the link below:

brainly.com/question/19943363

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5 0
2 years ago
What is the first step in nuclear fission?
lys-0071 [83]

Answer:

energy is released

Explanation:

8 0
3 years ago
2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'
Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

\displaystyle M_1V_1= M_2V_2

Where <em>M</em> represents molarity and <em>V</em> represents volume.

Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:

\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}

Convert this value to mL:
\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

8 0
2 years ago
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