Part B
DIRECTIONS: Identify the type of heat transfer depicted in each picture by arranging the
jumbled letters and answer the questions that follows.

Answer:

it is from the currents the movements cause

Explanation:

Answer:The greater the difference in pressure, the faster the winds. When a high- and a low-pressure air mass are far apart, air moves slowly from high to low pressure.  When a high- and a low-pressure air mass are far apart, air moves quickly from high to low pressure.

Explanation:

Answer:

Option C (1.2 L) is the correct option.

Explanation:

The given values are:

Pressure,

[tex]P_1=1.9 \ atm[/tex]

[tex]P_2=3.8 \ atm[/tex]

Volume,

[tex]V_1=2.3 \ L[/tex]

[tex]V_2=?[/tex]

As we know,

The gas equation is:

⇒  [tex]PV=nRT[/tex]

At constant temperature,

⇒  [tex]P_1 V_1=P_2 V_2[/tex]

or,

⇒      [tex]V_2=\frac{P_1 V_1}{P_2}[/tex]

By substituting the values, we get

⇒           [tex]=\frac{1.9\times 2.3}{3.8}[/tex]

⇒           [tex]=\frac{4.37}{3.8}[/tex]

⇒           [tex]=1.15 \ L[/tex]

or,

⇒           [tex]=1.2 \ L[/tex]

Answer:

100 L

Explanation:

Use Boyle's Law, p1v1 = p2v2.

p1 = 4 atm

v1 = 50L

p2 = 2 atm

v2 = ?

p1v1 = p2v2

(4 atm) (50L) = (2 atm) (v2)

v2 = (4 atm)(50L) / (2 atm) = 100 L

Answer:

II would say there is co2 so i would say 30.4

Explanation:

Hope this helps

Answer:

To find the relative formula mass (M r) of a compound, you add together the relative atomic mass values (A r values) for all the atoms in its formula.

Explanation: Find the M r of carbon monoxide, CO. Find the M r of sodium oxide, Na 2O. The relative formula mass of a substance, shown in grams, is called one mole of that substance.

Answer:

2C₄H₁₀(g) + 13O₂(g) —> 8CO₂(g) + 10H₂O(g)

Explanation:

The balanced equation for the reaction between butane and oxygen to produce carbon dioxide and water can be written as follow:

C₄H₁₀ + O₂ —> CO₂ + H₂O

Next, we shall balanced the equation. This is illustrated below:

C₄H₁₀ + O₂ —> CO₂ + H₂O

There are 4 atoms of C on the left side and 1 atom on the right side. It can be balance by writing 4 before CO₂ as shown below:

C₄H₁₀ + O₂ —> 4CO₂ + H₂O

There are 10 atoms of H on the left side and 2 atoms on the right side. It can be balance by writing 5 before H₂O as shown below:

C₄H₁₀ + O₂ —> 4CO₂ + 5H₂O

There are 2 atoms of O on the left side and a total of 13 atoms on the right side. It can be balance by writing 13/2 before O₂ as shown below:

C₄H₁₀ + 13/2O₂ —> 4CO₂ + 5H₂O

Multiply through by 2 to express in whole number ratio.

2C₄H₁₀(g) + 13O₂(g) —> 8CO₂(g) + 10H₂O(g)

Thus, the equation is balanced.

Answer: The amount of heat released when 233 grams of water at room temperature (25C) is frozen into ice at -15C is 17253.65 J.

Explanation:

As water converts into ice at [tex]0^{o}C[/tex]. So, the change in temperature is taking place from [tex]25^{o}C[/tex] to [tex]0^{o}C[/tex] and then from [tex]0^{o}C[/tex] to [tex]-15^{o}C[/tex].

So, amount of heat released by water is as follows.

[tex]q_{1} = m \times C \times (T_{2} - T_{1})\\= 233 g \times 4.18 J/g^{o}C \times (25 - 0)^{o}C\\= 24348.5 J[/tex]

m = mass of substance

C = specific heat of water = [tex]4.18 J/g^{o}C[/tex]

[tex]C'[/tex] = specific heat of ice = [tex]2.03 J/g^{o}C[/tex]

The amount of heat released by ice is as follows.

[tex]q_{2} = m \times C' \times (T'_{2} - T'_{1})\\= 233 g \times 2.03 J/g^{o}C \times (-15 - 0)^{o}C\\= -7094.85[/tex]

Now, total heat released during this change is as follows.

[tex]\Delta H = q_{1} + q_{2}\\= 24348.5 J + (-7094.85) J\\= 17253.65 J[/tex]

Thus, we can conclude that heat released when 233 grams of water at room temperature (25C) is frozen into ice at -15C is 17253.65 J.

Answer:

Explanation:

This is an example of a limiting reactant question, and is very common as a general chemistry problem.

We first see the balanced equation, that is:

2CuCl2+4KI→2CuI+4KCl+I2

We first need to find the limiting reactant

We see that 0.56 g of copper(II) chloride (CuCl2) reacts with 0.64 g of potassium iodide (KI) . So, let's convert those amounts into moles.

Copper(II) chloride has a molar mass of

134.45 g/mol . So in 0.56 g of copper(II) chloride, then there exist

0.56g134.45g/mol≈4.17⋅10−3 mol

Potassium iodide has a molar mass of

166 g/mol . So, in 0.64 g of potassium iodide, there exist

if it wrong i am sorry

Answer:

The arrangement of electrons within an atom is called electron configuration

If I help you, will you help other people?

Actually "this is the worst question i've ever seen use the internet bud" why don't you do that.

Answer:

0.702 /s

Explanation:

Rate constant at [tex][298 \mathrm{~K}, \mathrm{~K}_{1}=3.46 \times 10^{-2} \mathrm{~s}^{-1}[/tex]

Rate constant at [tex]350 \mathrm{~K}, \mathrm{~K}_{2}=?[/tex]

[tex]T_{1}=298 \mathrm{~K}[/tex]

[tex]T_{2}=350 \mathrm{~K}[/tex]

Activation energy, [tex]\mathrm{Ea}=50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}[/tex]

Use the following equation to calculate [tex]K_{2}$ at $350 \mathrm{~K}[/tex]

Use the following equation to calculate [tex]K_{2}$ at $350 \mathrm{~K}[/tex]

[tex]\ln \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right][/tex]

Therefore,

 [tex]\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{1}{298 \mathrm{~K}}-\frac{1}{350 \mathrm{~K}}\right][/tex]

[tex]\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} \times 350 \mathrm{~K}}\right][/tex]

[tex]\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=\mathrm{e}^{3.01}[/tex]

[tex]\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=20.3[/tex]

[tex]K_{2} &=20.3 \times 3.46 \times 10^{-2} \mathrm{~s}^{-1}[/tex]

[tex]&=0.702 \mathrm{~s}^{-1}[/tex]

hence, the rate constant at [tex]350 \mathrm{~K} [/tex] is 0.702[tex]\mathrm{~s}^{-1}[/tex]

Explanation:

asexual reproduction involves 1 parent and produces offspring that are genetically identical to each other and parent.

Sexual reproduction involves 2 parents and produces offspring that are genetically unique.

Answer:

D . It describes a compound being broken down into the elements from which it was formed.

Explanation:

A compound consists of two or more elements chemically combined together in a chemical reaction. The compound carbon monoxide is represented by the chemical symbol, CO. This means that it contains 1 atom of the carbon element and 1 atom of the Oxygen element.

This compound is formed when carbon is directly oxidized in a limited supply of oxygen. In the above reaction, this compound is disintegrated into its individual components, namely, carbon and oxygen.

Equation balance:

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

2Al(OH)3+3H2SO4

H=12

SO4=3

O=6

Al=2

Al2(SO4)3+6H2O

H=12

SO4=3

O=6

Al=2

Answer:

2Cu2O + C --> 4Cu + CO2

Explanation

The Law of conservation of Mass states You can not destroy or create math in a chemical reaction, meaning you must have the same amount at the beginning and the end. This equation keeps the same amount of atoms on both sides.

Answer: Yes, it is a balanced equation.

Explanation:

https://chemequations.com/en/?s=Cu2O+%2B+C+%3D+Cu+%2B+CO2

This is where I got the answer so you know.  Hope this helps :)

It's ammonia (NH3).....

Answer:

A molecule with one lone pair and three bond pair

- Ammonia

The nitrogen gas and the oxygen gas will both occupy 50 liters.

Let us recall form the kinetic theory of gases that a gas is in constant random motion. The gas does not have a specific volume but takes on the volume of the container since it will expand and fill the container.

The volume occupied by the two gases will be the same as the volume of the container. Hence, the nitrogen gas and the oxygen gas will both occupy 50 liters.

Learn more about volume of a gas: https://brainly.com/question/4519953

Answer:

B. The entire container because gases will expand to fill it.

Explanation:

I took to test, plus it directly says this in the lesson

Answer:

True

Explanation:

Answer:

true

Explanation:

source(s): trust me bro

Answer:

D

Explanation:

Because it has equal number of elements on each side of the arrow

Answer:

P2 = 4.76 atm

Explanation:

Given:

T1 = 235K. P1 = 3.80 atm

T2 = 298K. P2 = ?

Here, we use Gay-Lussac's law:

P1/T1 = P2/T2

Solving for P2, we get

P2 = (T2/T1)P1

= (298K/238K)(3.80 atm)

= 4.76 atm

Answer:

250cm³ de disolución se deben preparar

Explanation:

El porcentaje volumen volumen (%v/v) se define como el volumen de soluto -En este caso, H2O2- en 100cm³ de disolución.

Se desea preparar una solución que contenga 20cm³ de H2O2 por cada 100cm³

Como solamente se cuenta con 50cm³ de H2O2, el volumen que se debe preparar es:

50cm³ H2O2 * (100cm³ Disolución / 20cm³ H2O2) =

Answer:

A. 5.3 X 10^-2

Explanation:

I got 100 on PLATO

Explanation:

amount of baking soda = 2.01 g

---> 2.01 g NaHCO3 × (1 mol NaHCO3/84.007 g NaHCO3)

= 0.024 mol NaHCO3

mass of the product = 1.28 g

Possible reactions:

NaHCO3(s) ---> NaOH(s) + CO2(g)

0.024 mol NaHCO3 × (1 mol NaOH/1 mol NaHCO3)

= 0.024 mol NaOH × (39.997 g NaOH/1.mol NaOH)

= 0.96 g NaOH

2NaHCO3(s) ---> Na2O(s) + 2CO2(g) + H2O(g)

0.024 mol NaHCO3 × (1 mol Na2O/2 mol NaHCO3)

= 0.012 mol Na2O × (61.9789 g Na2O)

= 0.73 g Na2O

2NaHCO3(s) ---> Na2CO3(s) + CO2(g) + H2O(g)

0.024 mol NaHCO3 × (1 mol Na2CO3/2 mol NaHCO3)

= 0.012 mol Na2CO3 × (105.9888 g Na2CO3/1 mol Na2CO3)

= 1.27 g Na2CO3

Conclusion:

Based on the mass of the product in the crucible and comparing it with the predicted product mass in different reactions, I conclude that Na2CO3 is the most likely product in the crucible.

Answer:

4.17M

Explanation:

m1v1/m2v2=n1/n2

Answer:

6.1 M

Explanation:

Step 1: Calculate the moles of HNO₃ in the initial solution

A student has 475 mL (0.475 L) of a 4.5 M HNO₃ (4.5 mol HNO₃/ L) solution.

0.475 L × 4.5 mol HNO₃/ L = 2.1 mol HNO₃

He leaves the top of the container open, so water will evaporate, but the moles of HNO₃ will remain the same.

Step 2: Calculate the final concentration of the solution

2.1 moles of HNO₃ are in 347 mL (0.347 L) of solution.

[HNO₃] = 2.1 mol / 0.347 L = 6.1 M

Answer:

water vapor (gas) is becoming liquid