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2 years ago

Which of the following events could increase Earth's temperature?

Chemistry

2 answers:

Tcecarenko [31]2 years ago

5 0

I believe the answer Is B

ehidna [41]2 years ago

3 0

Answer:

The answer is C & A.

Hope this helped you out!^^

You might be interested in

If .758 moles of gas occupy a volume of 80.6L, how many moles will occupy a volume of 270.9L?

egoroff_w [7]

Answer:

n₂ = 2.55 mol

Explanation:

Given data:

Initial number of moles = 0.758 mol

Initial volume = 80.6 L

Final volume = 270.9 L

Final number of moles = ?

Solution:

Formula:

V₁/n₁ = V₂/n₂

V₁ = Initial volume

n₁ = initial number of moles

V₂ = Final volume

n₂ = Final number of moles

now we will put the values in formula.

V₁/n₁ = V₂/n₂

80.6 L / 0.758 mol = 270.9 L/ n₂

n₂ = 270.9 L× 0.758 mol / 80.6 L

n₂ = 205.34 L.mol /80.6 L

n₂ = 2.55 mol

4 0

2 years ago

What is the binding energy of a nucleus that has a mass defect of 5.81*10-^29 kg

IrinaVladis [17]

Answer:

Choice A: Approximately $5.23 \times 10^{-29}$ joules.

Explanation:

Apply the famous mass-energy equivalence equation to find the energy that correspond to the $\rm 5.81\times 10^{-29}$ kilograms of mass.

$E = m \cdot c^{2}$ ,

where

$E$ stands for energy,
$m$ stands for mass, and
$c$ is the speed of light in vacuum.

The speed of light in vacuum is a constant. However, finding the right units for this value can simplify the calculations a lot. What should be the unit of $c$ ?

The mass given is in the appropriate SI unit:

Mass is in kilograms.

Thus, proceed with the speed of light in SI units. The SI unit for speed is meters per second. For the speed of light, $c \approx \rm 3.00\times 10^{8}\;m\cdot s^{-1}$ .

Apply the mass-energy equivalence:

$\begin{aligned} E &= m \cdot c^{2} \\ &= \rm 5.81\times 10^{-29}\; kg \times {\left(3.00\times 10^{8}\; m\cdot s^{-1}\right)}^{2}\\ &\approx \rm 5.23\times 10^{-12}\;kg\cdot m^{2}\cdot s^{-2} \end{aligned}$ .

The unit of energy is not in joules. Don't be alerted. Consider the definition of a joule of energy. One joule is the work done on an object when a force of one newton acts on the object in the direction of the force through the distance of one meter. (English Wikipedia.)

$\rm 1\; J = 1\; N \times 1\; m$ .

However, a force of one newton is defined as the force required to accelerated an object with a mass of one kilogram (not gram) at a rate of one meter per second squared. (English Wikipedia.)

$\begin{aligned}\rm 1\; J &= \rm 1\; N \times 1\; m\\ & = \rm \left(1\; kg\times 1\; m\cdot s^{-2}\right)\times 1\; m\\ &= \rm 1\; kg \cdot m^{2}\cdot s^{-2}\end{aligned}$ .

In other words, the mass defect here is also $\rm 5.23\times 10^{-12}\; J$ .

7 0

3 years ago

Read 2 more answers

What is the significance of a gene?Immersive Reader

Ilia_Sergeevich [38]

Explanation:

A gene is the basic physical and functional unit of heredity. Genes are made up of DNA. Every person has two copies of each gene, one inherited from each parent. Most genes are the same in all people, but a small number of genes (less than 1 percent of the total) are slightly different between people. These small differences contribute to each person’s unique physical features.

4 0

2 years ago

3. What detrmines the identity of an atom?

Brrunno [24]

The number of protons in one atom of an element determines the atom's identity, and the number of electrons determines its electrical charge. The atomic number tells you the number of protons in one atom of an element. ... The atomic mass of an element is the total number of protons and neutrons in the nucleus of the atom.

5 0

2 years ago

Calculate Delta H in KJ for the following reactions using heats of formation:

lozanna [386]

Answer:

$\Delta H\textdegree = -2856.8\;\text{kJ}$ per mole reaction.

$\Delta H\textdegree = -22.3\;\text{kJ}$ per mole reaction.

Explanation:

What is the standard enthalpy of formation $\Delta H_f\textdegree{}$ of a substance? $\Delta H_f\textdegree{}$ the enthalpy change when one mole of the substance is formed from the most stable allotrope of its elements under standard conditions.

Naturally, $\Delta H_f\textdegree{} = 0$ for the most stable allotrope of each element under standard conditions. For example, oxygen $\text{O}_2$ (not ozone $\text{O}_3$ ) is the most stable allotrope of oxygen. Also, under STP $\text{O}_2$ is a gas. Forming $\text{O}_2\;(g)$ from itself does not involve any chemical or physical change. As a result, $\Delta H_f\textdegree{} = 0$ for $\text{O}_2\;(g)$ .

Look up standard enthalpy of formation $\Delta H_f\textdegree{}$ data for the rest of the species. In case one or more values are not available from your school, here are the published ones. Note the state symbols of the compounds (water/steam $\text{H}_2\text{O}$ in particular) and the sign of the enthalpy changes.

$\text{C}_2\text{H}_6\;(g)$ : $-84.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{CO}_2\;(g)$ : $-393.5\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{H}_2\text{O}\;{\bf (g)}$ : $-241.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}\;(s)$ : $-217.9\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}_2\;(s)$ : $-276.6\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{Pb}_3\text{O}_4\;(s)$ : $-734.7\;\text{kJ}\cdot\text{mol}^{-1}$

How to calculate the enthalpy change of a reaction $\Delta H_\text{rxn}$ (or simply $\Delta H$ from enthalpies of formation?

Multiply the enthalpy of formation of each product by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Reactants}))$ to show that this value takes the coefficients into account.
Multiply the enthalpy of formation of each reactant by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Products}))$ to show that this value takes the coefficient into account.
Change = Final - Initial. So is the case with enthalpy changes. $\Delta H_\text{rxn} = \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))$ .

For the first reaction:

$\Sigma (n\cdot \Delta_f(\text{Reactants})) = 4\times (-393.5) + 6\times (-241.8) = -3024.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\Sigma (n\cdot \Delta_f(\text{Products})) = 2\times (-84.0) + 7\times 0 = -168.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\begin{aligned}\Delta H_\text{rxn} &= \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))\\ &= (-3024.8\;\text{kJ}\cdot\text{mol}^{-1}) - (-168.0\;\text{kJ}\cdot\text{mol}^{-1})\\ &= -2856.8\;\text{kJ}\cdot\text{mol}^{-1} \end{aligned}$ .

Try these steps for the second reaction:

$\Delta H_\text{rxn} = -22.3\;\text{kJ}\cdot\text{mol}^{-1}$ .

6 0

3 years ago