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inna [77]
3 years ago
6

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

38.5 cm3 piece of wood out of his pile of materials, what would the mass be if that pile was all american white oak?
Chemistry
1 answer:
mixas84 [53]3 years ago
7 0

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

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the individual atom in the molecule

Explanation:

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I hope this answer will help you. Have a nice day !

5 0
2 years ago
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Answer: usuhduhduhdjhd

Explanation:

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

7 0
3 years ago
calculate δg o for each reaction using δg o f values: (a) h2(g) i2(s) → 2hi(g) 2.6 kj (b) mno2(s) 2co(g) → mn(s) 2co2(g) kj (c)
Reika [66]

(a)The change in Gibbs free energy for the reaction has been 2.6 kJ/mol.

(b) The change in Gibbs free energy for the reaction has been -49.3 kJ/mol.

(c) The change in Gibbs free energy for the reaction has been 91.38 kJ/mol.

6 0
1 year ago
PLEASE HELP I'LL GIVE YOU BRAINLIEST!!
In-s [12.5K]

Answer:

C

Explanation:

5 0
2 years ago
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