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Masteriza [31]
3 years ago
15

Which situation would lead to increased competition

Chemistry
1 answer:
I am Lyosha [343]3 years ago
8 0

I would say B because c and d would decrease competition and a would do the same, or just kill the ecosystem.

Hope this helps and don't forget to hit that heart :)

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How do atoms combine to form all of the diverse types of matter in our universe? Pls, Help with a good valid answer and explanat
german
I don’t understand this
3 0
3 years ago
According to the following reaction, how many grams of potassium phosphate will be formed upon the complete reaction of 29.6 gra
Ugo [173]

Answer:

There is 37.36 grams of K3PO4 produced

Explanation:

Step 1: Data given

Mass of H3PO4 = 29.6 grams

KOH is in excess

Molar mass of KOH = 56.11 g/mol

Molar mass of H3PO4 = 97.99 g/mol

Step 2: The balanced equation

3KOH(aq) + H3PO4(aq) ⇔ K3PO4(aq)+3H2O(l)

Step 3: Calculate mass of KOH

Mass KOH = mass KOH / molar mass KOH

Mass KOH = 29.6 grams / 56.11 g/mol

Mass KOH = 0.528 moles

Step 4: Calculate moles of K3PO4

Since KOH is the limiting reactant, We need 3 moles of KOH for each moles of H3PO4, to produce 1 mole of K3PO4 and 3 moles of H2O

For 0.528 moles of KOH we'll have 0.528/3 =  0.176 moles of K3PO4

Step 5: Calculate mass of K3PO4

Mass K3PO4 = moles K3PO4 * molar mass K3PO4

Mass K3PO4 = 0.176 moles * 212.27 g/mol

Mass K3PO4 = 37.36 grams

There is 37.36 grams of K3PO4 produced

8 0
3 years ago
Hydrogen peroxide can decompose to water and oxygen by the following reaction
mestny [16]
To begin calculating, there is one thing you need to remember :1 mole of H2O2=34.0148 g
Then we have 5.00g of H2O2=5/34.0148=0.146995
As you know decomposition of 2moles now has prodused <span>196kj
So, </span><span>q is made due </span>0.146995 moles of H2O2=(-196/2)*0.146995=-14.40551Kj
I'm sure it will help.
6 0
3 years ago
Name the covalent compound:<br><br> I8P
SOVA2 [1]

There are certain rules to follow when naming covalent compounds. But first, let us look at the definition of Covalent Compounds.

<h3>What are Covalent Compounds?</h3>

When covalent bonds aid the creation of a molecule, in which the atoms have at least one similar pair of valence electrons, a covalent compound is said to have been formed.

A very common example is water (H₂O)

<h3>How are Covalent Compounds named?</h3>

To name a covalent compound, simply list the first element in the formula using the name of the element, then name the second element by adding the suffix "ide" to the stem of the second element's name.

If there is only one atom in the molecule of the first element, then no prefix should be added.

It is to be noted that if the second element in the compound is oxygen, then we should say:

  • monox<em>ide</em> instead of monoox<em>ide</em> and
  • triox<em>ide</em> instead of trox<em>ide</em>, all depending on how many atoms that are involved.

See the attached for the prefixes related to the various number of atoms in the compounds.

It is to be noted that the covalent compound to be named here is not stated hence the general answer.

Learn more about naming covalent compounds at:
brainly.com/question/9841865

Download txt
6 0
2 years ago
A common laboratory reaction is the neutralization of an acid with a base. When 31.8 mL of 0.500 M HCl at 25.0°C is added to 68.
lord [1]

Answer:

The correct answer to the following question will be "90.6 kJ/mol".

Explanation:

The total reactant solution will be:

(31.8 \ mL+68.9 \ mL)\times 1.07\ g/mL = 107.74 \ g

The produced energy will be:

=4.18 \ J/(gK)\times 107.74 \ g\times (28.2-25.0)K

=450.35\times 3.2

=1441.12 \ J

The reaction will be:

⇒  HCl+NaOH \rightarrow NaCl+H_{2}O

Going to look at just the amounts of reactions with the same concentrations, we notice that they're really comparable.  

Therefore, the moles generated by NaCl will indeed be:

=  (\frac{31.8}{1000} \ L)\times (0.500 \ M \ HCl/L)\times \frac{1 \ mol \ NaCl}{1 \ mol \ HCl}

=  0.0318\times 0.500

=  0.0159 \ mole  \ of \ NaCl

Now,

=  \frac{1441.12 \ J}{0.0159 \ moles \ NaCl}

=  906364.7

=  90.6 \ KJ/mol \ NaCl

7 0
3 years ago
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