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If 1,000 mL = 1 L, which of the following are possible conversion factors for liters and milliliters? Check all that apply.

Fudgin [204]

Answer:

Two conversion factors:

$1=\dfrac{1L}{1,000ml}\\ \\ \\ 1=\dfrac{1,000mL}{1L}$

Explanation:

You can create two possible conversion factors, one to convert from mL to L, and one to convert from L to mL

a) From mL to L

To convert mL to L you need to multiply by a conversion factor that has mL on the denominator and L in the numerator.

Your starting point is: $1,000mL=1L$

Then, divide both sides by 1,000mL (this will be on the denominator of the fraction);

$1,000mL=1L\\\\ \\\dfrac{1,000ml}{1,000mL}=\dfrac{1L}{1,000mL}\\ \\ \\ 1=\dfrac{1L}{1,000mL}$

b) From L to mL

Divide both sides by 1 L:

$1,000mL=1L\\\\ \\\dfrac{1,000ml}{1L}=\dfrac{1L}{1L}\\ \\ \\ 1=\dfrac{1,000mL}{1L}$

7 0

3 years ago

Read 2 more answers

Assuming constant pressure, rank these reactions from most energy released by the system to most energy absorbed by the system,

givi [52]

Answer: The order from the Most energy released to most Energy Absorbed Is given as 2---> 4--->,3-->---> 1

B)-61.9 kJ

Explanation:

The change in the internal energy of a system is positive if the reaction absorbs energy and negative if the reaction releases energy. For a system to cause an increase in volume, it must have very high energy built up to be released.

1. Surroundings get colder and the system decreases in volume. Here, the surrounding absorbs energy resulting in positive ΔE

2. Surroundings get hotter and the system expands in volume. Here energy is released causing the system to be negative

3. Surroundings get hotter and the system decreases in volume. Although there is a decreased volume, the system is negative because it releases energy

4. Surroundings get hotter and the system does not change in volume. System is negative because it releases energy even thgoygh there is no change in volume

Therefore the order from the Most energy released to most Energy Absorbed Is given as 2---> 4--->,3-->---> 1

b) Using

ΔE = q+ w from 1st law of thermodynamics

ΔE= ΔH - P ΔV

gIven

ΔH = -75.0KJ

volume= A change from 5.0L TO 2.0L = Final volume - initial volume = 2-5= -3.00L

P= 43.0atm

ΔE= ΔH - P ΔV

P ΔV = 43 atm x -3 = -129L.atm

We first convert L-atm to Joules.

1 L-atm = 101.325 Joules.

129L.atm = 129 x 101.325 = - 13071 J

to KJ becomes

13071/1000 = - 13.071KJ

Recall ΔE= ΔH - P ΔV and putting values

ΔE = -75.0 - (-13.071 KJ)= -75.0 kJ + 13.071 kJ = -61.9 kJ

8 0

3 years ago

CO2+2H2O=H3O^++HCO3^-

bekas [8.4K]

Answer:

Bronsted Lowry Theory

$CO_2$ is the acid

$H_2 O$ is the base

$H_3 O^+$ is the conjugate acid and

$HCO_3^-$ is the conjugate base

Explanation:

Bronsted Lowry Theory:

An acid is a substance that can donate one or more protons

A base is a substance which can accept one or more protons

Hydrogen atom which is neutral (No Charge) contains 1 positive proton, 1 negative electron and 0 neutral neutron.

Thus Hydrogen atom has no Charge and it is neutral.

When an hydrogen atom loses an electron, Hydrogen ion is formed, which will contain 1 positive proton and 0 negative electron and 0 neutral neutron.

Thus Hydrogen ion has a positive charge.

Hydrogen ion is also called as a proton since it has only 1 proton in it.

Hydrogen ion in water that is,

$H^++H_2O\Rightarrow H_3O^+$

$H_3O^+$ is called as Hydronium ion.

Acid loses H+ (proton) to form conjugate Base

Base gains H+ (proton) to form conjugate Acid

For example

Let us consider the example given in the question

$CO_2+2H_2 O>H_3 O^+ +HCO_3^-$ can be written as (removed 1 $H_2 O$ from both the sides )

$CO_2+H_2 O>H^++HCO_3^-$ or

$CO_2+H_2 O > H_2 CO_3$ reversing the equation

$H_2 CO_3+H_2 O>H_3 O^+ +HCO_3^-$

$H_2 CO_3$ is the acid which donates $H^+$ to form $HCO_3^-$

$H_2 O$ is the base which gains $H^+$ to form $H_3 O^+$

$HCO_3^-$ is the conjugate base and $H_3 O^+$ is the conjugate acid

So

$CO_2$ is the acid

$H_2 O$ is the base

$H_3 O^+$ is the conjugate acid and

$HCO_3^-$ is the conjugate base

(Answer)

5 0

2 years ago

What does multivalent mean???

salantis [7]

In the context of multivalent ions, it is when it has multiple oxidative states.

4 0

2 years ago

1) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data

earnstyle [38]

Answer:

1) The absolute zero temperature is -272.74 °C

2) The absolute zero temperature is -269.91°C

Explanation:

1) The specific volume of air at the given temperatures are;

At -85°C, v = 1/(1.877) = 0.533 dm³/g, the pressure =

At 0°C, v = 1/1.294 ≈ 0.773 dm³/g

At 100°C, v = 1/0.946 ≈ 1.057 dm³/g

We therefore have;

v₁ = 0.533 T₁ = 188.15 K

v₂ = 0.773 T₂ = 273.15 K

v₃ = 1.057 T₃ = 373.15 K

The slope of the graph formed by the above data is therefore given as follows;

m = (1.057 - 0.533)/(373.15 - 188.5) = 0.00284 dm³/K

The equation is therefor;

v - 0.533 = 0.00284×(T - 188.15)

v = 0.00284×T - 0.534 + 0.533 = 0.00284×T - 0.001167

v = 0.00284×T - 0.001167

Therefore, when the temperature, at absolute 0, we have;

v = 0

Which gives;

0 = 0.00284×T - 0.001167

0.00284×T = 0.001167

T = 0.001167/0.00284 ≈ 0.411 K

Which is 0.411 -273.15 ≈ -272.74 °C

The absolute zero temperature is -272.74 °C

2) The given volume of the gas = 20.00 dm³ at 0°C and 1.000 atm

The slope of the volume temperature graph at constant pressure = 0.0741 dm³/(°C)

The temperature is converted to Kelvin temperature, in order to apply Charles law as follows;

0°C = 0 + 273.15 K = 273.15 K

Therefore, the equation of the graph can be presented as follows;

v - 20.00 = 0.0741 × (T - 273.15)

Which gives;

v = 0.0741·T - 0.0741 ×(273.15) + 20

v = 0.0741·T - 20.2404125 + 20

v = 0.0741·T - 0.240415

Therefore at absolute 0, v = 0, we have;

0 = 0.0741·T - 0.240415

0.0741·T = 0.240415

T = 0.240415/0.0741 = 3.2445 K

The temperature in degrees is therefore;

3.2445 K - 273.15 ≈ -269.91°C

The absolute zero temperature is therefore, -269.91°C

3 0

3 years ago