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3 years ago

An object has a mass of 441 g and a volume

Chemistry

1 answer:

Ratling [72]3 years ago

6 0

Answer:

<h3>The answer is 4.41 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

$density = \frac{441}{10} \\$

We have the final answer as

Hope this helps you

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What is veces??????????????????????????????????

tangare [24]

Veces is Spanish word for times.Hope I helped.

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3 years ago

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For <em>hydrogen </em>we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for <em>nitrogen</em>:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that <u>3 moles of hydrogen</u> react with <u>1 mol of nitrogen</u>, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have <u>3.47 moles of hydrogen</u> and we need <u>7.50 moles</u> to react with all the mass of nitrogen, the <em>limiting reactant</em> is <em>hydrogen</em>.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that <u>3 moles of hydrogen</u> produces <u>2 moles of ammonia</u>, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$

Hence, hydrogen would produce <u>2.31 moles of ammonia</u>.

Therefore, hydrogen is the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

I hope it helps you!

6 0

3 years ago

Consider the following chemical equilibrium:

jok3333 [9.3K]

Answer:

Kc = [CH₄] / [H₂]²

Kp = [CH₄] / [H₂]² * (0.082*T)^-1

Explanation:

Equilibrium constant, Kc, is defined as the ratio of the concentrations of the products over the reactants. Also, each concentration of product of reactant is powered to its coefficient.

<em>Pure solids and liquids are not taken into account in an equilibrium</em>

Thus, for the reaction:

C(s)+ 2H₂(g) ⇌ CH₄(g)

Equilibrium constant is:

Now, using the formula:

Kp = Kc* (RT)^Δn

<em>Where R is gas constant (0.082atmL/molK), T is the temperature of the reaction and Δn is difference in coefficients of gas products - coefficients of gas reactants (1 - 2= -1)</em>

Replacing:

<em />

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3 years ago

How are the electron structures of boron (B) and aluminum (Al) similar?

andrezito [222]

The electron structures of boron and aluminum are similar because they share the same group, therefore they have the same amount of valence electron.

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3 years ago

Read 2 more answers

What is the frequency of a photon having an energy of 4.91 × 10–17 ? (c = 3.00 × 108 m/s, h = 6.63 × 10–34 J · s)

AlexFokin [52]

Answer:

The frequency of the photon is 7.41*10¹⁶ Hz

Explanation:

Planck states that light is made up of photons, whose energy is directly proportional to the frequency of radiation, according to a constant of proportionality, h, which is called Planck's constant. This is expressed by:

E = h*v

where E is the energy, h the Planck constant (whose value is 6.63*10⁻³⁴ J.s) and v the frequency (Hz or s⁻¹).

So the frequency will be:

$v=\frac{E}{h}$

Being E= 4.91*10⁻¹⁷ J and replacing:

$v=\frac{4.91*10^{-17} J}{6.63*10^{-34} J.s}$

You can get:

v= 7.41*10¹⁶ $\frac{1}{s}$ = 7.41*10¹⁶ Hz

<u><em>The frequency of the photon is 7.41*10¹⁶ Hz</em></u>

4 0

3 years ago