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3 years ago

If all the matter in the universe was birthed seconds after the Big Bang, then what was likely triggered the bang?

Chemistry

1 answer:

ArbitrLikvidat [17]3 years ago

8 0

Answer:

Welp God made everything so that means the big bang never happened so that means techniqually the answer to your question is nothing ig.

hope this helps

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What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

3 years ago

For the following aqueous reaction, complete and balance the molecular equation and write a net iconic equatio, making sure to i

pav-90 [236]

Answer:

Balance molecular equation:

K2CO3(aq) + Sr(NO3)2(aq) → SrCO3(s) + 2KNO3(aq)

Net ionic equation:

CO3∧-2(aq) + Sr∧+2(aq) → SrCO3(s)

Explanation:

Potassium carbonate = K2CO3

Strontium nitrate = Sr(NO3)2

Chemical equation:

K2CO3 + Sr(NO3)2 → SrCO3 + KNO3

Balance chemical equation with physical states:

K2CO3(aq) + Sr(NO3)2(aq) → SrCO3(s) + 2KNO3(aq)

Ionic equation:

2K+(aq) + CO3∧-2(aq) + Sr∧+2(aq) + 2NO∧-3(aq) → SrCO3(s) + 2K+(aq) + 2NO∧-3(aq)

Net ionic equation:

CO3∧-2(aq) + Sr∧+2(aq) → SrCO3(s)

2K+ and 2NO∧-3 ions are spectator ions that's way these are not written in net ionic equation.

Spectator ions:

These are the ions that are present same on both side of chemical reaction and does not effect the equilibrium.

6 0

3 years ago

What is speed? If you walk 3 kilometers in 30 minutes, what is your average speed in kilometers per hour? Compare and contrast i

Mrrafil [7]

Answer:

See explanation

Explanation:

distance = 3 kilometres

time = 30 minutes or 0.5 hours

speed = distance/time = 3/0.5 = 6 Kmhr-1

Average speed is given by total distance covered divided by the total time taken while the instantaneous speed is the speed of an object at a particular instant or moment in time. Average speed refers to the whole journey while instantaneous speed refers to a particular point or instant in the journey.

distance= x

time = 3 hours

speed = 50 miles per hour

speed = distance/time

distance = speed * time

distance = 3 * 50

distance = 150 miles

3 0

3 years ago

What is a common hazard of commercial fishing

timurjin [86]

like stuff getting stuck to the fishermen hooks , net etc

4 0

4 years ago

Read 2 more answers

Flag this question question 8 10 pts use the δh°f and δh°rxn information provided to calculate δh°f for if: δh°f (kj/mol) if7(g)

GarryVolchara [31]

$\Delta H\textdegree{}_f(\text{IF} \; (g)} = -95 \; \text{kJ} \cdot \text{mol}^{-1}$

Explanation

$\text{IF}_7 \; (g) + \text{I}_2 \; (s) \to \text{IF}_5 \; (g) + 2\; \text{IF} \; (g)$

$\Delta H\textdegree{}_\text{rxn} = -89\; \text{kJ} \cdot \text{mol}^{-1}$
$\Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) = -941 \; \text{kJ} \cdot \text{mol}^{-1}$ (from the question)
$\Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) = -840 \; \text{kJ} \cdot \text{mol}^{-1}$ (from the question)

As an the most stable allotrope under standard conditions, $\Delta H\textdegree{}_f (\text{I}_2) = 0\; \text{kJ} \cdot \text{mol}^{-1}$

By definition,

$\Delta H\textdegree{}_\text{rxn} = \Delta H\textdegree{}_f (\text{all products}) - \Delta H\textdegree{}_f (\text{all reactants})$

$\Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + 2 \; \Delta H\textdegree{}_f (\text{IF} \; (g) ) - \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) - \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) \\ = \Delta H\textdegree{}_{\text{rxn}}$

$\begin{array}{ccc} \Delta H\textdegree{}_f (\text{IF} \; (g) )& = & 1/2\; ( \Delta H\textdegree{}_{\text{rxn}} - \Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) + \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) )\\ & = & 1/2 \; (-89 - (-840) + (-941))}\\ & = & - 95 \; \text{kJ} \cdot \text{mol}^{-1} \end{array}$

Note, that iodine on the reactant side is stated as a gas in the equation given in the question whereas under standard conditions it is expected to be under the solid state; the $\Delta H\textdegree{} _f$ given in the question seemingly corresponds to the one in which the reactant iodine exists as a solid rather than as a gas. Evaluating the last expression using data from an external source

$\Delta H\textdegree{}_f (\text{I}_2 \; (g) ) = \Delta H\textdegree{}_f(\text{I}_2 \; (s)) + \Delta H\textdegree{}_{\text{sublimation}}(\text{I}_2) = 62.42 \; \text{kJ} \cdot \text{mol}^{-1}$ (Cox, Wagman, et al., 1984)

... yields $\Delta H\textdegree{}_f (\text{IF} \; (g) ) \approx -64 \; \text{kJ}\cdot \text{mol}^{-1}$ , which deviates significantly from the experimental value of $-94.76 \; \text{kJ}\cdot \text{mol}^{-1}$ (Chase, 1998.) It is thus assumed that the $\Delta H\textdegree{}_\text{rxn}$ value provided requires a reaction with $\text{I}_2 \; (s)$ rather than $\text{I}_2 \; (g)$ as a reactant.

3 0

4 years ago