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SpyIntel [72]
2 years ago
9

PLEASEEEE ANSWERR 15 POUNTS

Chemistry
2 answers:
LekaFEV [45]2 years ago
7 0

Answer:

c

Explanation:

Lina20 [59]2 years ago
5 0

Answer:

240

Explanation:

15 POUNTS 240

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When 25.0 grams of solid potassium hydroxide (KOH, molar mass = 56.1 g/mol) is dissolved in 100.0 grams of water, the solution i
Maru [420]

Answer:

They gave you the equation; Cp=,

just plug everything in! You’ve seen this; I have long ago, but we had different units. Sorry, but it’s right there! Go get it!

Explanation:

4 0
3 years ago
Write a balanced chemical equation for the standard formation reaction of solid sodium hydrogen carbonate Write a balanced chemi
4vir4ik [10]

Answer:

Na(s) + C(s, graphite) + 1/2 H₂(g) + 3/2 O₂(g) → NaHCO₃(s)

Explanation:

The standard formation reaction is the synthesis of 1 mole of a substance from its elements in their most stables forms under standard conditions. The balanced chemical equation is:

Na(s) + C(s, graphite) + 1/2 H₂(g) + 3/2 O₂(g) → NaHCO₃(s)

8 0
3 years ago
What is the precipitation reaction of sodium nitrate and lead (II)?
Vadim26 [7]
The precipitation reaction of sodium nitrate and lead (II) chloride would not happen. No reaction will happen between these substances. Hope this answers the question. Have a nice day.
8 0
3 years ago
The reaction below was carried out in an acidic solution. Upper I minus, plus upper I upper O minus subscript 3 right arrow uppe
Ganezh [65]

Answer:

It has been balanced by using the half-reaction method.

Explanation:

I- and IO3- gives I2

We divide the reaction into two half-reactions

(2 I- >> I2 + 2e-) x5 ( oxidation  : I goes from -1 to 0 )

2 IO3- + 12H+ + 10e- >> I2 + 6H2O ( reduction : I goes from +5 to 0 )

10 I- >> 5I2 + 10e-

2IO3- + 12H+ + 10e- >> I2 + 6H2O

-----------------------------------------------------

10 I- + 2IO3- + 12H+ >> 6I2 + 6H2O

To get the smallest numbers we divide by 2 :

5 I- + IO3- + 6H+ >> 3I2 + 3H2O

5 0
3 years ago
Read 2 more answers
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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