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grandymaker [24]
4 years ago
7

Convert the boiling temperature at gold, 2966 degree centigrate , into degrees fahrenheit and kelvin.

Chemistry
1 answer:
nexus9112 [7]4 years ago
7 0

Answer:

Boiling point of gold is 1630 ^{0}\textrm{F} or 3239.15 K

Explanation:

We know, T(^{0}\textrm{F})=1.8T(^{0}\textrm{C})+32

Here T(^{0}\textrm{C}) and T(^{0}\textrm{F}) are temperature in degree centigrate and temperature in degree farhenite respectively.

So, 2966^{0}\textrm{C}=\frac{2966-32}{1.8}^{0}\textrm{F}

                                            = 1630 ^{0}\textrm{F}

We know, 1^{0}\textrm{C}=273.15 K

So, 2966 ^{0}\textrm{C} = (2966+273.15) K = 3239.15 K

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How many moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas?
tino4ka555 [31]

Taking into account the reaction stoichiometry, 340.0 moles  of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CO₂ + 4 H₄  → CH₄ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • CO₂: 1 mole
  • H₄: 4 moles  
  • CH₄: 1 mole
  • H₂O: 2 moles

<h3>Moles of CH₄ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

moles of CH_{4} =\frac{85.1 moles of CO_{2}x4 moles of CH_{4} }{1 moles of CO_{2}}

<u><em>moles of CH₄= 340.4 moles</em></u>

Then, 340.0 moles  of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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6 0
2 years ago
2 HCl + 1 Zn -&gt; 1 H2 + ZnCl2 what type of reaction is this
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Single replacement reaction

Explanation:

3 0
3 years ago
In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to
labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of NH_3 = 100 g

Molar mass of NH_3 = 27 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate moles of NH_3.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles

The given balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the given reaction, we conclude that

2 moles of NH_3 produced from 1 mole of N_2

3.7 moles of NH_3 produced from \frac{1mole}{2mole}\times 3.7mole=1.85moles of N_2

Now we have to calculate the mass of N_2.

Mass of N_2 = Moles of N_2 × Molar mass of N_2

Mass of N_2 = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0
3 years ago
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