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Virty [35]
3 years ago
12

QUESTION 5 Match the symbol with the type of symbiotic relationship it represents

Chemistry
2 answers:
Kobotan [32]3 years ago
8 0
A:4
B:3
C:1
D:2
There u go hope this helped you
marshall27 [118]3 years ago
6 0

Answer:

A : 1

B : 3

C : 4

D : 2

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When 8.0 grams of sodium hydroxide is dissolved in sufficient water to make 400. mL of solution, what is the concentration of th
nata0808 [166]

Answer:

The concentration of this sodiumhydroxide solutions is 0.50 M

Explanation:

Step 1: Data given

Mass of sodium hydroxide (NaOh) = 8.0 grams

Molar mass of sodium hydroxide = 40.0 g/mol

Volume water = 400 mL  = 0.400 L

Step 2: Calculate moles NaOH

Moles NaOH = mass NaOH / molar mass NaOH

Moles NaOH = 8.0 grams / 40.0 g/mol

Moles NaOh = 0.20 moles

Step 3: Calculate concentration of the solution

Concentration solution = moles NaOH / volume water

Concentration solution = 0.20 moles / 0.400 L

Concentration solution = 0.50 M

The concentration of this sodiumhydroxide solutions is 0.50 M

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3 years ago
What is the molarity of a 500.0 ml solution containing 0.75 moles of solute?
wariber [46]
M=mol/L
M=.75/.5
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3 years ago
Which three of the five items would be best to include in a safety contract?
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A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
AysviL [449]

Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, q=mc\Delta T .

where, q is the heat energy, m is mass, c is specific heat and \Delta T is change in temperature.

Combined mass of calorimeter and water is 250.0 g and the specific heat is \frac{1.035cal}{g.^0C} .

\Delta T  for calorimeter and water (combined) = 11.08 - 10.00 = 1.08 degree C

\Delta T  for metal = 11.08 - 45.00 = -33.92 degree C

let's plug in the values in the above equation and calculate heat gained by combined system.

q=250.0g*\frac{1.035cal}{g.^0C}*1.08^0C

q = 279.45 cal

So, the heat lost by the metal is 279.45 cal.


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3 years ago
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What are two ways that nonmetallic minerals can be used?
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Idk if i'm too late but the answer should be A.

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