What is true for the energy involved in an explosion?

1 answer:

6 0

The correct answer is this one: "The amount of energy before and after the explosion depends on the type of reaction." The energy involved in an explosion is that the amount of energy before and after the explosion depends on the type of reaction, how strong and how weak; how destructive or less destructive.

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One cup of fresh orange juice contains 115 mg of ascorbic acid (vitamin c, c6h8o6). given that one cup

kodGreya [7K]

The question is incomplete. Complete question is read as:
'One cup of fresh orange juice contains 115 mg of ascorbic acid (vitamin C, C6H8O6). Given that one cup = 218.0 mL calculate the molarity of vitamin C in organic juice.'
..........................................................................................................................
Answer:
Given: weight of solute (ascorbic acid) = 115 mg = 0.115 g
Volume of solution = 218.0 mL = 0.218 L
Molecular weight of ascorbic acid = 176.12 g/mol.

Now, Molarity = $\frac{\text{weight of solute(g)}}{\text{Molecular weight X Vol. of solution(l)}}$
= $\frac{\text{0.115}}{\text{176.12 X 0.218}}$
 = 0.002995 mol/dm3

Answer: Molarity of solution = 0.002995 mol/dm3

4 0

3 years ago

Which of these was a characteristic of the rutherford model of the atom, but not the thomson model?

andrew-mc [135]

Thompson didnt have a nucleus (since rutherford discovered it)

7 0

3 years ago

Which of the following steps correctly converts 3 moles of fluorine to an equivalent number of particles of fluorine?

Mrrafil [7]

Multiply 3 by the atomic mass of fluorine, forgive me if I’m wrong

4 0

3 years ago

Need help on #129. Please help!

MrRissso [65]

The percentage yield is 72.8 %.

Step 1. Calculate the mass of Br₂

Mass of Br₂ = 20.0 mL Br₂ × (3.10 g Br₂/1 mL Br₂) = 62.00 g Br₂

Step 2. Calculate the theoretical yield

M_r: 159.81 266.69

2Al + 3Br₂ → 2AlBr₃

Moles of Br₂ = 62.00 g Br₂ × (1 mol Br₂/(159.81 g Br₂) = 0.3880 mol Br₂

Moles of AlBr₃ = 0.3880 mol Br₂ × (2 mol AlBr₃/(3 mol Br₂) = 0.2586 mol AlBr₃

Theor. yield of AlBr₃ = 0.2586 mol AlBr₃ × 266.99 g AlBr₃)/(1 mol AlBr₃)

= 69.05 g AlCl₃

Step 3. Calculate the percentage yield

% yield = (actual yield/theoretical yield) × 100 % = (50.3 g/69.05 g) × 100 %

= 72.8 %

5 0

3 years ago

A student lifts a box of books that weighs 350 N. The box is lifted 4.0 m. How much work does the student do on the box?

stepan [7]

Explanation:

F=350N

d=4m

W=F*d

=350*4

=1400 NM

5 0

4 years ago