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IceJOKER [234]
3 years ago
12

Why do you think there are so many types of fungi on the rainforest floor?

Chemistry
2 answers:
trapecia [35]3 years ago
7 0
Mold grows faster in the dark, where increased moisture means a higher relative humidity.
Tcecarenko [31]3 years ago
4 0

Answer:

Fungi can grow in dark, humid places that get little sunlight.

Explanation:

You might be interested in
Please give me the answer please
zepelin [54]

Answer:

A. 30cm³

Explanation:

Based on the chemical reaction:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

<em>1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂</em>

<em />

To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:

<em>Moles CaCO₃ -Molar mass: 100.09g/mol-</em>

1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles

<em>Moles HCl:</em>

50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles

For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:

2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.

The moles produced of CO₂ are:

2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂

Using PV = nRT

<em>Where P is pressure = 1atm assuming STP</em>

<em>V volume in L</em>

<em>n moles = 1.25x10⁻³ moles CO₂</em>

<em>R gas constant = 0.082atmL/molK</em>

<em>T = 273.15K at STP</em>

<em />

V = nRT / P

1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V

0.028L = V

28cm³ = V

As 28cm³ ≈ 30cm³

Right option is:

<h3>A. 30cm³</h3>

5 0
3 years ago
If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course),
garri49 [273]

Answer:

0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH

Explanation:

The pKa's of phosphoric acid are:

H₃PO₄/H₂PO₄⁻ = 2.1

H₂PO₄⁻/HPO₄²⁻ = 7.2

HPO₄²⁻/PO₄³⁻ = 12.0

To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻

To have a 50mM solution of phosphoures we need:

2L * (0.050mol / L) = 0.10 moles of H₃PO₄

0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4

To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:

H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺

H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺

Using H-H equation we can find the amount of NaOH added:

pH = pKa + log [A⁻] / [HA] <em>(1)</em>

<em>Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻</em>

<em>pH = 7.40</em>

<em>pKa = 7.20</em>

[A-] + [HA] = 0.10moles <em>(2)</em>

Replacing (2) in (1):

7.40 = 7.20 + log 0.10mol - [HA] / [HA]

0.2 = log 0.10mol - [HA] / [HA]

1.5849 = 0.10mol - [HA] / [HA]

1.5849 [HA] = 0.10mol - [HA]

2.5849[HA] = 0.10mol

[HA] = 0.0387 moles = H₂PO₄⁻ moles

That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles

The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles

And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles

Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH

Then, you need to dilute both solutions to 2.00L with distilled water.

4 0
3 years ago
What kind of changes occur between atoms when substances undergo chemical reactions?
KatRina [158]
The atoms undergo chemical changes.

New substances may be a product of the reaction.
8 0
3 years ago
How many grams of iron(III) oxide would be required to make 187 g of iron
enyata [817]

Answer:267 g

Explanation:

4 0
3 years ago
I’m struggling with dimensional analysis in chemistry! Will someone please help me with .74 Kcal/min to cal/sec ? Explain and gi
valina [46]

Answer:

            12.33 cal/sec

Explanation:

As we know,

                          1 Kcal  =  1000 cal

So,

                     0.74 Kcal  =  X cal

Solving for X,

                      X =  (0.74 Kcal × 1000 cal) ÷ 1 Kcal

                      X =  740 cal

Also we know that,

                      1 Minute  =  60 Seconds

Therefore, in order to derive cal/sec unit replace 0.74 Kcal by 740 cal and 1 min by 60 sec in given unit as,

                                       = 740 cal / 60 sec

                                       = 12.33 cal/sec

3 0
3 years ago
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