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Katarina [22]
3 years ago
15

How many grams of hydrogen are needed to react with 1.40 g of nitrogen to produce ammonia?

Chemistry
1 answer:
vazorg [7]3 years ago
8 0

Answer:

Write a balanced chemical reaction:

N2 + 3H2 ==> 2NH3

Looking at the mole ratios in this balanced equation you can see it takes 3 moles H2 to make 2 moles NH3.  So, next calculate the moles of NH3 represented by 1.80 g and then convert to moles of H2 needed:

moles of NH3 = 1.80 g x 1 mole/17 g = 0.106 moles NH3

Moles H2 needed = 0.106 moles NH3 x 3 moles H2/2 moles NH3 = 0.159 moles H2 needed

Grams H2 needed = 0.159 moles x 2 g/mole = 0.318 grams H2 needed

Explanation:

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Radioactive decay can be described by the following equation where is the original amount of the substance, is the amount of the
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Iron remains = 17.49 mg

Explanation:

Half life of iron -55 = 2.737 years (Source)

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{2.737}\ year^{-1}

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[A_0] = 32.2 mg

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

So,  

[A_t]=32.2\times e^{-0.2533\times 2.41}\ mg

[A_t]=32.2\times e^{-0.610453}\ mg

[A_t]=17.49\ mg

<u>Iron remains = 17.49 mg</u>

8 0
3 years ago
How many electrons does carbon need to gain to obtain a noble gas electron configuration
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The absorbance (????)(A) of a solution is defined as ????=log10(????0????) A=log10⁡(I0I) where ????0I0 is the incident‑light int
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Answer:

The absorbance of the myoglobin solution across a 1 cm path is 0.84.

Explanation:

Beer-Lambert's law :

Formula used :

A=\epsilon \times c\times l

A=\log \frac{I_o}{I}

\log \frac{I_o}{I}=\epsilon \times c\times l

where,

A = absorbance of solution

c = concentration of solution

\epsilon = Molar absorption coefficient

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I_o = incident light

I = transmitted light

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l = 1 cm, c = 1 mg/mL ,\epsilon = 15,000 M^{-1}cm^{-1}

Molar mass of myoglobin = 17.8 kDa = 17.8 kg/mol=17800 g/mol

(1 Da = 1 g/mol)

c = 1 mg /mL = {1mg /mL}{\text{Molar mass of myoglobin}}

c = \frac{1 mg/mL}{ 17800 g/mol} = 5.6179\times 10^{-5} mol/L

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A= 15,000 M^{-1}cm^{-1}\times 5.6179\times 10^{-5} mol/L\times 1 cm

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The absorbance of the myoglobin solution across a 1 cm path is 0.84.

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4 years ago
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