Determine the molar mass of Cao

PLEASE HELP!!!! 20 MINS LEFT IM DYING PLS HELP

Zepler [3.9K]

Answer:

Kc = 0.20

Explanation:

N₂O₄ ⇄ 2NO₂

moles 5.3mol 2.3mol

Vol 5L 5L

Molarity 5.3/5M 2.3/5M

= 1.06M = 0.46M

Kc = [NO₂]²/[N₂O₄] = (0.46)²/(1.06) = 0.1996 ≅ 0.20

6 0

3 years ago

8. True or False: There are some exceptions to the rule.... organisms such as a protist called a

ankoles [38]

There are some exceptions to the rule organisms such as a protist called a euglena can be both heterotrophic and autotrophic. This is a true statement.

Explanation:

Euglena is a large genus of unicellular protists: they have both plant and animal characteristics
Photoautotrophs include protists that have chloroplasts, such as Spirogyra. Heterotrophs get their energy by consuming other organisms. Other protists can get their energy both from photosynthesis and from external energy sources

All live in water and move by means of a flag ellum. This is an animal characteristic. Most have chloroplasts, which are characteristic of algae and plants
Euglena is photosynthetic in the presence of sunlight i.e autotrophic, when deprived of sunlight they behave like heterotrophs by predating on other smaller organisms.
Most species of Euglena have photosynthesizing chloroplasts within the body of the cell, which enable them to feed by autotrophy, like plants. They can also take nourishment heterotrophically, like animals.

6 0

3 years ago

The density of air under ordinary conditions at 25°C is 1.19 g/L. How many kilograms of air are in a room that measures 10.0 ft

Lunna [17]

Answer: The mass of air present in the room is 37.068 kg

Explanation : Given,

Length of the room = 10.0 ft

Breadth of the room = 11.0 ft

Height of the room = 10.0 ft

To calculate the volume of the room by using the formula of volume of cuboid, we use the equation:

$V=lbh$

where,

V = volume of the room

l = length of the room

b = breadth of the room

h = height of of the room

Putting values in above equation, we get:

$V=10.0ft\times 11.0ft\times 10.0ft=1100ft^3=31148.53L$

Conversion used : $1ft^3=28.3168L$

Now we have to calculate the mass of air in the room.

$Density=\frac{Mass}{Volume}$

$1.19g/L=\frac{Mass}{31148.53L}$

$Mass=37066.7507g=37.068kg$

Conversion used : (1 kg = 1000 g)

Therefore, the mass of air present in the room is 37.068 kg

3 0

3 years ago

Read 2 more answers

Production of Jam from Crushed Fruit in Two Stages. In a process producing jam (C1), crushed fruit containing 14 wt % soluble so

Andrej [43]

Answer:

The total kg from the mixer: $m=2222.5 kg$

Evaporated water: $m_{water ev.}=791.1 kg$

Jam produced: $m_{jam}=1431.4 kg$

Explanation:

Step by step:

1) Kg of mixture from the mixer

It says that crushed fruit is added to the mixer with the sugar and pectin.

The fruit added is 1000 kg

The sugar added is 1.22 kg sugar/1 kg crushed fruit:

$m_{sugar}=1000 kg fruit * \frac{1.22 kg sugar}{1 kg fruit}=1220 kg sugar$

The pectin added is 0.0025 kg pectin/1 kg crushed fruit:

$m_{pectin}=1000 kg fruit * \frac{0.0025 kg pectin}{1 kg fruit}=2.5 kg pectin$

The total kg from the mixer:

$m=1000 kg fruit + 2.5 kg pectin + 1220 kg sugar=2222.5 kg$

2) Evaporated water

To calculate the evaporated water it's important to have in mind that the mix goes from a concentration of 14 wt% to 67 wt%. This difference is because of the evaporated water. So:

Initial: 1000 kg of fruit with 14 wt% of solids

$m_{solids}=1000 kg *0.14=140 kg$