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Nuetrik [128]
3 years ago
15

Which statements describe an interaction between the biosphere and the atmosphere related to photosynthesis? Select 2 correct ch

oices.
During photosynthesis, plant roots take in water from soil.
During photosynthesis, plants take in carbon dioxide from the air.
Energy stored in the soil during photosynthesis is released into the air.
During photosynthesis, plants make food and release oxygen into the air.
Energy stored in plants through photosynthesis is transferred to humans who eat them.
Chemistry
2 answers:
nika2105 [10]3 years ago
8 0

Answer:b

Explanation:

coldgirl [10]3 years ago
7 0

Answer:

It is the second one and the fourth one

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A science student observes that a dissolvable antacid tablet acts faster in relieving stomach pain when taken with warm water th
LekaFEV [45]

Answer:the medication would not last as long

Explanation:

Because of the desolving of the tablet so quickly it would

4 0
4 years ago
Read 2 more answers
A galvanic cell at a temperature of 42 degrees Celcius is powered by the following redox reaction:
seropon [69]

<u>Answer:</u> The cell voltage of the given reaction is 1.86 V

<u>Explanation:</u>

The given chemical equation follows:

3Cu^{2+}(aq.)+2Al\rightarrow 2Al^{3+}(aq.)+2Au(s)

<u>Oxidation half reaction:</u> Al(aq.)\rightarrow Al^{3+}(aq.)+3e^-;E^o_{Al^{3+}/Al}=1.66V       ( × 2)

<u>Reduction half reaction:</u> Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.16V       ( × 3)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.16-(-1.66)=1.82V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +1.82 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol Kl

T = temperature = 42^oC=[42+273]K=315K

F = Faraday's constant = 96500

[Al^{3+}]=1.63M

[Cu^{2+}]=3.43M

Putting values in above equation, we get:

E_{cell}=1.82-\frac{2.303\times 8.314\times 315}{2\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})\\\\E_{cell}=1.86V

Hence, the cell voltage of the given reaction is 1.86 V

4 0
3 years ago
4. The heat of
Reika [66]

It is 2 KJ

It is because the formula is :

q=m∆Hf

where m is the mass and ∆Hf is the heat fusion.

So, we get:

q=80×25

q=2000J or 2 KJ

8 0
3 years ago
When 4.088 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 13.82 grams of CO2 and 2.829 grams of H
yarga [219]

Answer:

Empirical formula: CH

Molecular formula: C₆H₆

Explanation:

Based on the combustion of a hydrocarbon, the moles of CO₂ = Moles of Carbon in the hydrocarbon and the moles of H₂O = 1/2 moles of hydrogen in the hydrocarbon.

The empirical formula is the simplest whole number of atoms present in a molecule. With the moles of C and H we can find empirical formula:

<em>Moles C -Molar mass CO₂ = 44.01g/mol-:</em>

13.82g * (1mol / 44.01g) = 0.314 moles C

<em>Moles H -Molar mass H₂O = 18.01g/mol-:</em>

2.829g H₂O * (1mol / 18.01g) = 0.157 moles H₂O * (2mol H / 1mol H₂O) = 0.314 moles of H

The ratio of moles H: moles C:

0.314 moles / 0.314 moles = 1

That means empirical formula is:

<h3>CH</h3><h3 />

With the molecular weight and empirical formula we can find the molecular formula:

Molar mass CH = 12.01g/mol+1.01g/mol = 13.02g/mol

As the molecular weight of the molecule is 78.11amu = 78.11g/mol, there are:

78.11g/mol / 13.02g/mol = 6 times the empirical formula in the molecular formula

That means molecular formula is:

<h3>C₆H₆</h3>
4 0
3 years ago
How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)
zhannawk [14.2K]

Answer:

Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

4 0
3 years ago
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