What change in energy occurs when you burn coal or petroleum?

On a coordinate plane, which single transformation could map point A (3,-4) onto point A' (1.5,-2)? *

Ksenya-84 [330]

Answer: A dilation with rule: $(x,y)\to(\dfrac12x,\dfrac12y)$

Explanation:

A dilation is a non-rigid transformation that creates an image that is the same shape as the original but has a different size.

It uses a scale factor k such that

$(x,y)\to(kx,ky)$

(x,y)= coordinates of original figure

(kx,ky) = corresponding coordinate in the image.

To transform: A (3,-4) onto point A' (1.5,-2).

Using scale factor k= $\dfrac12$ , we have

$A(3,-4)\to A'(\dfrac12\times3, \dfrac12\times-4)=A'(1.5,-2)$

Required rule: $(x,y)\to(\dfrac12x,\dfrac12y)$

3 0

3 years ago

a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what

jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in liters.

$n$ - Molar quantity, measured in moles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

$\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}$ (2)

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$V_{1}$ , $V_{2}$ - Initial and final volume, measured in liters.

$T_{1}$ , $T_{2}$ - Initial and final temperature, measured in Kelvin.

If we know that $P_{1} = 121\,kPa$ , $P_{2} = 202\,kPa$ , $V_{1} = 2.7\,L$ , $T_{1} = 288\,K$ and $T_{2} = 303\,K$ , the final volume of the gas is:

$V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}$

$V_{2} = 1.702\,L$

The gas will occupy a volume of 1.702 liters.

6 0

3 years ago

A 4,000 mL solution of AgNO3 contains 17.00 g of solute in water. Calculate the molar concentration of the solution.

blsea [12.9K]

The molarity is moles/liters.

First, convert 4,000 mL to L:

4000 mL --> 4 L

Now, you must convert the 17 g of solute to moles by dividing the number of grams by the molar mass. The molar mass of AgNO3 is 169.87 g/mol:

17 / 169.87 = .1

Now that you have both the number of moles and the liters, plug them into the initial equation of moles/liters:

.1/4 = .025

3 0

3 years ago

Read 2 more answers

What would be the new volume if the temperature on 800 mL of gas is changed

dolphi86 [110]

I think you divide something

8 0

3 years ago

1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in

FrozenT [24]

Answer:

1) 10.0 moles of NO

2) 25 moles of NaCl

3) 1200 moles of CO2

4) 1.03 moles of MgO

5) 0.72 moles H2

6) 1041.15 grams BaCl2

7) 9.55 grams MgO

8) 45.5 grams Au

9 )14.93 grams AlCl3

Explanation:

1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in the following composition reaction?

N2 + O2 → 2NO

For 1 mol N2 we need 1 mol O2 to produce 2 moles of NO

For 5.0 moles of N2 we need 5.0 moles of O2 to produce 10.0 moles of NO

2. The neutralization of an acid with a base is a double replacement reaction in which a salt and water are formed. If you start with 25 moles of HCl and neutralize it with NaOH how many moles of NaCl will be formed?

HCl + NaOH → NaCl + H2O

For 1 mol HCl we need 1 mol NaOH to produce 1 mol of NaCl and 1 mol H2O

For 25 moles of HCl we need 25 moles of NaOH to produce 25 moles of NaCl and 25 moles of H2O

3. A car burns gasoline (octane – C8H18) with oxygen. If you drive to Salt Lake and burn 150 moles of octane how many moles of carbon dioxide are you producing?

2C8H18 + 25O2 → 16CO2 + 18H2O

For 2 moles of octane we need 25 moles of O2 to produce 16 moles of CO2 and 18 moles of H2O

For 150 moles of octane we need 25*75 = 1875 moles of O2

To produce 16*75 = 1200 moles of CO2 and 18*75= 1350 moles

4. If 25 gram of magnesium combines with oxygen in a composition reaction, how many moles of magnesium oxide will be formed?

2Mg + O2 → 2MgO

Moles of Mg = 25.0 g/24.3 g/mol = 1.03 moles

For 2 moles Mg we need 1 mol O2 to produce 2 moles MgO

For 1.03 moles Mg we'll have 1.03 moles of MgO

5. . Lithium reacts with water in a single replacement reaction. How many moles of hydrogen gas a produced by 10 grams of lithium?

2Li + 2H2O → 2LiOH + H2

Moles Li = 10.0 grams/ 6.94 g/mol = 1.44 moles

For 2 moles Li we need 2 mole H2O to produce 2 moles LiOH and 1 mol H2

For 1.44 moles Li we need 1.44 moles H2O to produce 1.44 moles H2O and 0.72 moles H2

6. Barium chloride reacts with sodium sulfate in a double replacement reaction. How many grams of barium chloride are required to react with 5 moles of sodium sulfate?

BaCl2 + Na2SO4 → BaSO4 + 2NaCl

For 1 mol of BaCl2 we need 1 mol of Na2SO4 to produce 1 mol of BaSO4 and 2 moles NaCl

For 5 moles Na2SO4 we need 5 moles BaCl2

mass BaCl2 = 5 moles * 208.23 g/mol = 1041.15 grams BaCl2

7. Magnesium carbonate when heated decomposes to form magnesium oxide and carbon dioxide. How many grams of magnesium oxide will be formed if 20 grams of magnesium carbonate are heated?

MgCO3 → MgO + CO2

Moles MgCO3 = 20.0 grams / 84.31 g/mol

Moles MgCO3 = 0.237 moles

For 1 mol MgCO3 we'll have 1 mol MgO and 1 mol CO2

For 0.237 moles MgCO3 we'll have 0.237 moles MgO and 0.237 moles CO2

Mass MgO = 0.237 moles * 40.30 g/mol = 9.55 grams MgO

8. If 70 grams of gold III chloride decomposes into its elements, how many grams of gold will be produced?

2AuCl3 → 2Au + 3Cl2

Moles AuCl3 = 70 grams / 303.33 g/mol = 0.231 moles

For 2 moles AuCl3 we'll have 2 moles gold and 3 moles Cl2

For 0.231 moles AuCl3 we'll have 0.231 moles gold

Mass of gold = 0.231 moles * 196.97 g/mol = 45.5 grams Au

9. Chlorine is more reactive element than bromine, thus chlorine will replace bromine in compound through a single replacement reaction. If 30 grams of aluminum bromide react with chlorine in this fashion how many grams of aluminum chloride will be formed?

2AlBr3 + 3Cl2 → 2AlCl3 + 3Br2

Moles AlBr3 = 30 g /266.69 g/mol = 0.112 moles

For 2 moles AlBr2 we need 3 moles Cl2 to produce 2 moles AlCl3 and 3 moles Br2

For 0.112 moles AlBr3 we need 3/2 * 0.112 = 0.168 moles of Cl2

To produce 0.112 moles of AlCl3 and 0.168 moles of Br2

Mass AlCl3 = 0.112 moles * 133.34 g/mol = 14.93 grams AlCl3

8 0

4 years ago