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solniwko [45]
3 years ago
8

A sample of metal had a mass of 17.29g and a volume of 5.68mL what is the density of this metal

Chemistry
1 answer:
Gennadij [26K]3 years ago
8 0
Density = mass/volume

<span>17.29g / 5.68mL = 3.044g/mL</span>
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Strontium sulfate becomes less soluble in an aqueous solution when sodium sulfator is added because
horsena [70]

Answer:

The addition of sulfate ions shifts equilibrium to the left.

Explanation:

Hello!

In this case, according to the following ionization of strontium sulfate:

SrSO_3(s)\rightleftharpoons Sr^{2+}+SO_4^{2-}

It is evidenced that when sodium sulfate is added, sulfate, SO_4^{2+} is actually added in to the solution, which causes the equilibrium to shift leftwards according to the Le Ch athelier's principle. Thus, the answer in this case would be:

The addition of sulfate ions shifts equilibrium to the left.

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7 0
3 years ago
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What do all four of these types of molecules have in common?
nalin [4]
D. They all contain carbon as an important part of their structure.
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3 years ago
if the lightbulb receives 100 J of electrical energy, and gives off 75 energy, how much heat (thermal energy away from the light
oksano4ka [1.4K]

Answer:

Amount of heat energy released by light bulb = 25 joules

Explanation:

Given:

Energy receive by light bulb = 100 Joules

Energy released by light bulb as light energy = 75 Joules

Find:

Amount of heat energy released by light bulb

Computation:

We know that, energy is neither be created nor destroys

So,

Using Law of conservation of energy

Energy receive by light bulb = Energy released by light bulb as light energy + Amount of heat energy released by light bulb

100 = 75 + Amount of heat energy released by light bulb

Amount of heat energy released by light bulb = 100 - 75

Amount of heat energy released by light bulb = 25 joules

8 0
3 years ago
What element has the smallest atomic radius
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4 0
4 years ago
Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an
Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g    [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0
3 years ago
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