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3 years ago

Dry ice, or solid cabon dioxide, has the molecular formula of CO2. How should you classify it?

1 answer:

6 0

You should classify it as co2 solid as CO2(s) and that small left comes to the bottom right hand side and dry ice is a form of liquid carbon dioxide so u should write it as CO2(l) and that L comes to the bottom right hand sise

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How many molecules are in 4.2 miles of water?

viktelen [127]

Answer:

2.5284 x 10^24

4 0

3 years ago

What is the length of the shaded grey area? write your answer to the correct number of significant digits. The unit is cm

alukav5142 [94]

Answer:

1.37cm

Explanation:

It's less than 1.4cm but more than 1.3cm. It's also more than 1.35cm so I guess the best answer would be 1.37cm or round up to 1.4cm

8 0

2 years ago

Different kinds of wood have different densities. The density of american white oak wood is generally 0.77 g/cm3. If Jim grabs a

mixas84 [53]

Answer:

m= 29.645 g

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Given data:

density of wood = 0.77 g/cm³

volume= 38.5 cm³

mass= ?

Solution:

d= m/v

m= d × v

m= 0.77 g/cm³× 38.5 cm³

m= 29.645 g

7 0

3 years ago

When cyclohexene is treated with KMnO4, H2O, the syn-1,2-diol is produced. What reaction occurs when benzene is similarly treate

Ulleksa [173]

Answer:

No reaction is observed

Explanation:

The benzene ring is aromatic. Being an aromatic ring, the benzene ring is remarkably stable to all reactions that destroy the aromatic ring.

Alkenes are oxidized to alkanols in the presence of KMnO4 but this reaction does not occur with benzene. However, substituted benzenes having hydrogen atoms attached to the substituent carbon atom can be oxidized to the corresponding carboxylic acid.

7 0

3 years ago

When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°

dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0

2 years ago