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zhannawk [14.2K]
3 years ago
13

Read the following statement:

Chemistry
1 answer:
Whitepunk [10]3 years ago
4 0

Answer:

D

Explanation:

the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of an object

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Avagadros Law..<br>Definition please..​
Nikitich [7]

Answer:

"Avogadro's law is an experimental gas law relating the volume of a gas to the amount of substance of gas present. The law is a specific case of the ideal gas law. A modern statement is: Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."

8 0
3 years ago
Read 2 more answers
A certain solid has a density of 8.0 g/cm3. if 4.0 g of this solid are poured into 4.00 ml of water, which drawing below most cl
Alexandra [31]
Answer is: <span>the volume of water after the solid is added</span> is 4.5 ml.
d(gold) = 8.0 g/cm³; density of gold.
m(gold) = 4 g; mass of gold.
V(gold) = m(gold) ÷ d(gold); volume of gold.
V(gold) = 4 g ÷ 8 g/cm³.
V(gold) = 0.5 cm³ = 0.5 ml.
V(water) = 4.00 ml = 4.00 cm³.
V(flask) = V(gold) + V(water).
V(flask) = 0.5 cm³ + 4 cm³.V = 4.5 cm³.
4 0
3 years ago
How many moles are there in 35.2g of Gold?
ki77a [65]

Explanation:

196.96655

maybe

5 0
3 years ago
What does pressure measure
omeli [17]

Answer:

C

Explanation:

Pressure it says it in the name its the force of colliding particles

6 0
4 years ago
Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which
Drupady [299]

Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride =  0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be:= \dfrac{5 \ kg }{1590 \ kg/m^3}

= 0.0031 m³

Suppose \beta is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT

In ( \dfrac{V_2}{V_1})  = \beta (T_2-T_1)

V_2 = V_1 \times exp (\beta (T_2-T_1))

V_2 = 0.0031 \ m^3  \times exp  (1.2 \times 10^{-3} \times 20)

V_2 = 0.003175 \ m^3

However; the workdone = -PdV

W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)

W = - 7.6 J

The heat energy Q = Δ h

Q = mC_p(T_2-T_1)

Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ

3 0
3 years ago
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