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Step2247 [10]
3 years ago
6

Give two examples of amphibians. describe one similarity and difference between amphibians and fish

Chemistry
1 answer:
Nesterboy [21]3 years ago
7 0
Stop being a lil mark buster 
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What is a chemical property of the fruit shown above? A. The fruit can decompose. B. The fruit has a mass of 250 grams. C. The f
saw5 [17]

i think the best answer for this is B)


6 0
3 years ago
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Hello! I have a green pigment called the chlorophyll. I also absorb solar
vodka [1.7K]

Answer:

flower

Explanation:

4 0
2 years ago
How much energy in joules is released when 18.5 grams of copper cools from 285 °C<br> down to 45 °C
nadya68 [22]

Answer: 1709.4 Joules

Explanation:

The quantity of Heat Energy (Q) released on cooling a heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since Q = ?

M = 18.5 grams

Recall that the specific heat capacity of copper C = 0.385 J/g.C

Φ = 285°C - 45°C = 240°C

Then, Q = MCΦ

Q = 18.5grams x 0.385 J/g.C x 240°C

Q = 1709.4 Joules

Thus, 1709.4 Joules is released when copper is cooled.

5 0
3 years ago
Write and balance molecular equations for the following reactions between aqueous solutions. You will need to decide on the form
musickatia [10]

Answer:

This is the balanced equation:

Pb(NO₃)₂ (aq) + 2NaI (aq) → 2NaNO₃ (aq)  +  PbI₂ (s) ↓    

Explanation:

This are the reactants:

PbNO₃

NaI

Iodide can react to Pb²⁺ to make a solid compound.

4 0
3 years ago
What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0
3 years ago
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