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3 years ago

What is the mole-to-mole ratio between N2 and H2?

1 answer:

5 0

Remember, look at the coefficients in the balanced equation! Here are three mole ratios:
1 mole N2 / 3 moles H2.
3 moles H2 / 2 moles NH3.

You might be interested in

C3H8+3O2 = 3CO2+4H2O what is the enthalpy combustion

dezoksy [38]

Answer:

$\Delta H_{comb}=2043.85kJ/mol$

Explanation:

Hello there!

In this case, according to the given chemical reaction, it possible for us to set up the expression for the calculation of the enthalpy change as shown below:

$\Delta H_r=-\Delta H_{comb}=3\Delta _fH_{CO_2}+4\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-3\Delta _fH_{O_2}$

Thus, given the values of the enthalpies of formation on the attached file, we obtain: $-\Delta H_{comb}=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-103.85kJ/mol)-3(0kJ/mol)\\\\-\Delta H_{comb}=-2043.85kJ/mol\\\\\Delta H_{comb}=2043.85kJ/mol$

Best regards!

8 0

2 years ago

8. Which of the following is TRUE about limiting and excess reactants?

Ivahew [28]

B. The limiting reactant determines the max amount of product that can be formed

5 0

2 years ago

What is the predicted change in the boiling point of water when 1.50 g of

dezoksy [38]

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

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$\textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}$

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The elevation in boiling point is a phenomenon in which there is increase in boiling point in solution, when the particular type of solute is added to pure solvent.

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$\sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}$

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Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

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'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

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'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

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To find molality, we have to divide no. of moles of solute by weight of solution

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While first we need to no. of moles

$\sf \implies no. \: of \: moles = \frac{weight \: of \: solute}{molar \: mass \: of \: solute} \\ \\ \implies \sf no. \: of \: moles = \frac{1.5}{208.23} \\ \\ \sf \implies no. \: of \: moles = 0.0072$

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Now, we will find molality

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$\sf \hookrightarrow molality = \frac{no.\: of \: moles}{weight \: of \: solution} \\ \\ \sf \hookrightarrow molality = \frac{0.072}{1.5} \\ \\ \sf \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}$

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$\textsf{ \large{ \underline{Now substituting the required values}}}$

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$\sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}$

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Henceforth, the change in boiling point is 0.00735°C.

7 0

1 year ago

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an ex

Kay [80]

The question is incomplete. The complete question is :

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an explanation that tells how much NaOH needs to be added to a beaker of HCl to cause the color to change. Your explanation can be something like: The color change will occur when [some amount] of NaOH is added because the color change occurs when [some condition]. The goal for your explanation is that it describes the outcome of this example, but can also be used to predict the outcome of other examples of this phenomenon. Here's an example explanation: The color of the solution will change when 40 ml of NaOH is added to a beaker of HCl because the color always changes when 40ml of base is added. Although this explanation works for this example, it probably won't work in examples where the flask contains a different amount of HCl, such as 30ml. Try to make an explanation that accurately predicts the outcome of other versions of this phenomenon.

Solution :

Consider the equation of the reaction between NaOH and $HCl$

NaOH (aq) + HCl (aq) → NaCl(aq) + $H_2O (l)$

The above equation tells us that $1 \text{mole}$ of $NaOH$ reacts with $1 \text{mole}$ of $HCl$ .

So at the equivalence point, the moles of NaOH added = moles of $HCl$ present.

If the volume of the $HCl$ taken = $V_1$ mL and the conc. of $HCl$ = $M_1$ mole/L

The volume of NaOH added up to the color change = $V_2 \text{ and conc of NaOH = M}_2$ mole/L

Moles of $HCl$ taken = $V_1 \ mL \times M_1 \ mol/100 \ mL = V_2M_2 \times 10^{-3}$ moles.

The color change will occur when the moles of NaOH added is equal to the moles of $HCl$ taken.

Thus when $V_1 M_1 \times 10^{-3} = V_2M_2 \times 10^{-3}$

or when $V_1M_1 = V_2M_2$

or $V_2=\frac{V_1M_1}{M_2}$ mL of NaOH added, we observe the color change.

Where $V_1, M_1$ are the volume and molarity of the $HCl$ taken.

$M_2$ is the molarity of NaOH added.

When both the NaOH and $HCl$ are of the same concentrations, i.e. if $M_1=M_2$ , then $V_2=V_1$

Or the 40 mL of $HCl$ will need 40 mL of NaOH for a color change and

30 mL of $HCl$ would need 30 mL of NaOH for the color change (provided the concentration $M_1=M_2$ )

7 0

2 years ago

11. The back part of the cerebrum deals with hearing True False

Sunny_sXe [5.5K]

Answer:

The answer is: True

5 0

2 years ago