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vlada-n [284]
2 years ago
10

What is the compound name of O2CI4?

Chemistry
1 answer:
Dafna1 [17]2 years ago
6 0

Answer:

Dioxide tetrachloride

Explanation:

Di meaning 2 as in o2 and tetra the Greek numerical for 4 attached to chlorine in a group it is Chloride.

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A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3
IRINA_888 [86]

<em>The mass of a gold bar  = 1.424 x 10⁴ gram</em>

<em></em>

<h3><em>Further Explanation</em></h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

<em>Known variable</em>

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

<em>Asked</em>

the mass of a gold bar

<em>Answer</em>

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram

<h3><em>Learn more </em></h3>

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Keywords: density, mass, volume, a gold bar

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BigorU [14]
Hope this can help you

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One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 8.00 liters against a constant external pressure
Leya [2.2K]

Answer : The work done on the surroundings is, 709.1 Joules.

Explanation :

The formula used for isothermally irreversible expansion is :

w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)

where,

w = work done

p_{ext} = external pressure = 1.00 atm

V_1 = initial volume of gas = 1.00 L

V_2 = final volume of gas = 8.00 L

Now put all the given values in the above formula, we get :

w=-p_{ext}(V_2-V_1)

w=-(1.00atm)\times (8.00-1.00)L

w=-7L.atm=-7\times 101.3J=-709.1J

The work done by the system on the surroundings are, 709.1 Joules. In this, the negative sign indicates the work is done by the system on the surroundings.

Therefore, the work done on the surroundings is, 709.1 Joules.

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