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JulsSmile [24]
3 years ago
11

Increase food supplies in an ecosystem decrease competition because the competing organsims would

Chemistry
1 answer:
irga5000 [103]3 years ago
6 0
The competing organisms would eventually outnumber the competition
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What is the ph of the benzoic acid solution prior to adding sodium benzoate?
Reika [66]
The acid dissociation constant of benzoic acid is 6.5 x 10^-5. Therefore, the pH of the benzoic acid solution prior to adding sodium benzoate is:

pH = -log[Ka]
pH = -log (6.5 x 10^-5)
pH = 4.19

The pH of the benzoic acid solution is 4.19 which is acidic, but a weak acid. 

7 0
3 years ago
What are the 7 methods of separating mixtures?​
Gennadij [26K]

Methods Of Separating Mixtures

Handpicking.

Threshing.

Winnowing.

Sieving.

Evaporation.

Distillation.

Filtration or Sedimentation.

Separating Funnel.

3 0
3 years ago
Read 2 more answers
8
love history [14]

Answer : The molar concentration of sucrose in the tea is, 0.0549 M

Explanation : Given,

Mass of sucrose = 3.765 g

Volume of solution = 0.200 L

Molar mass of sucrose  = 342.3 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of sucrose}}{\text{Molar mass of sucrose}\times \text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{3.765g}{342.3g/mole\times 0.200L}=0.0549mole/L=0.0549M

Therefore, the molar concentration of sucrose in the tea is, 0.0549 M

7 0
3 years ago
A lead ball is added to a graduated cylinder containing 50.6 ml of water, causing the level of the water to increase to 93.0 mL.
Kamila [148]

42.4 ml is the volume in milliliters of the lead ball if a lead ball is added to a graduated cylinder containing 50.6 ml of water.

<h3>What is a graduated cylinder?</h3>

A tall narrow container with a volume scale is used especially for measuring liquids.

The graduated cylinder contains water

mL is a volume unit.

Water volume = 50.6 ml

The lead ball caused an increase in volume from 50.6 ml to 93.0 mL.

The new volume is the lead ball volume plus the original water volume :

Final volume = Vlead ball+ Water original volume

93.0 mL= V_(lead ball) +50.6 ml

V_(lead ball) = 93.0 mL - 50.6 ml

V_(lead ball) = 42.4 ml

Hence, 42.4 ml is the volume in milliliters of the lead ball.

Learn more about the graduated cylinder  here:

brainly.com/question/13386106

#SPJ1

4 0
2 years ago
calculate the mass required to prepare 2.5 L of 1.0 M NaOH solution. Given that the molar mass for NaOH is 40 g/mol.
Helen [10]

Answer:

The required mass to prepare 2.5 L of 1.0 M NaOH solution is 100 g

Explanation:

We do this by preparing the equation:

Mass = concentration (mol/L) x volume (L) x Molar mass

Mass = 1.0 M x 2.5 L x 40 g/mol

Mass = 100 g

3 0
3 years ago
Read 2 more answers
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