If you have 10g of Fe and 20 g of oxygen, how much product will you have?

A strand of hair is approximately 50 μm thick. If the diameter of an atom is 1 × 10-10 m, how many atoms thick is a strand of ha

Kazeer [188]

Thickness of strand hair = 50 μm

Diameter of an atom = $1 \times 10^{-10} m$

Now, with the help of conversion factors, convert thickness in micrometer into meter i.e. 1 micrometer = $10^{-6}m$

So, thickness of strand hair =50 μm= 50 \times 10^{-6} m

Therefore,

number of atoms thick is a strand of hair = $50 \times 10^{-6}m\times \frac{1 atom}{1 \times 10^{-10}m}$

= $50\times 10^{4}atoms$

Hence, number of atoms is $50\times 10^{4}atoms$ thick is a strand of hair.

6 0

3 years ago

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Megan prepares a pitcher of lemonade by adding a quarter cup of granular sugar to the mixture. Which action should she take so t

Licemer1 [7]

Can someone please help me with my chem review

3 0

3 years ago

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9.86 x 10²⁸ O-atoms require what volume (L) N₂O₂ at STP?

jolli1 [7]

First, we use avogadro's number to convert atoms into moles. Then, relate the number of moles from elemental to the compound. Lastly, we use conditions at STP to calculate the volume. We do as follows:

<span>9.86 x 10²⁸ O-atoms ( 1 mol / 6.022x10^23 atoms O) ( 1 mol N2O2 / 2 mol O ) ( 22.4 L / 1 mol ) = 1833809.37 L needed</span>

8 0

2 years ago

H-1

bonufazy [111]

Answer = B = Neutrons and Mass Number

Isotopes are defined as those atoms which have same atomic number but different atomic masses.

Atomic mass is basically the number of protons and neutrons present in an atom.

Atomic number is the number of protons present in an atom.

So, in isotopes the number of protons are same but the number of neutrons vary due to which atomic masses also vary.
In given three isotopes, all have same number of protons but different number of neutrons.
i.e.
H-1 = 1 P + 0 N = 1 u (Proton)
H-2 = 1 P + 1 N = 2 u (Deuterium)
H-3 = 1 P + 2 N = 3 u (Tritium)

Hence, it is clear that the number after H shows a change in number of neutrons and mass number.

7 0

3 years ago

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g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

Gala2k [10]

$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

$\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}$

$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

5 0

3 years ago