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Archy [21]
3 years ago
10

Which of the following is not an example of a molecule A. Mn     B. KOH    C. O₃    D. H₂S

Chemistry
2 answers:
Pavel [41]3 years ago
5 0

D. is your final answer!!!! :)))))

goldenfox [79]3 years ago
3 0

look at the chemical tables.  but i believe it is A

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A 2.5 mg sample of magnesium powder is ignited with 2 mg oxygen in a sealed container. All of the magnesium is consumed and 4.15
lys-0071 [83]

Answer:

Total mass of the reactant = 2+2.5 =4.5 mg

Total mass of product = 4.15 mg

therefore, mass of unreacted oxygen = 4.50-4.15 = 0.35 g

7 0
4 years ago
The variable that a scientist observes to change while conducting an experiment is called the ___ variable.
Nimfa-mama [501]

Great Question!

The Answer Would Be "B" The "RESPONDING" Variable

4 0
3 years ago
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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
4 years ago
What can be concluded about the atom from knowing that oxygen 18 has an atomic number of 8?
Vinvika [58]
<span>The atomic number of a neutral atom is equal to the number of protons and the number of electrons of the atom. The atomic weight meanwhile is equal to the sum of the number of protons and number of neutrons.Hence we can say that the number of protons is 8 and the number of neutrons is 10. </span>
7 0
3 years ago
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A 10g vial must have 45ml of diluent added to it. it has a powder volume of 5ml. how many mgr will be in each ml of the final so
Semenov [28]

The solution has a concentration 20 mgr in each mL of the final solution.

To solve this problem, we need to know about concentration. The concentration formula can be defined as how much the mass per unit volume is. It can be written as

M = m/V

where M is concentration, m is mass of solute, V is the total volume of solution.

From the text we know that :

m = 10g

vsolvent = 45mL

vsolute = 5 mL

find the total volume (V)

V = vsolvent + vsolute

V = 45 + 5

V = 50mL

Then, find the concentration

M = m/V

M = 10gr / 50 mL

M = 1000 mgr / 50mL

M = 20 mgr / mL

Hence, the solution has a concentration 20 mgr in each mL of the final solution.

Find more on concentration at: brainly.com/question/17206790

#SPJ4

6 0
2 years ago
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