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ryzh [129]
3 years ago
5

A roller coaster approaches a hill,

Chemistry
1 answer:
Lesechka [4]3 years ago
8 0

Answer:

the height of the hill is 30 feet tall

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What is the molarity of a solution made by dissolving 20.0 g of H3PO4 in 50.0 mL of solution
Rom4ik [11]
(20*1000)÷(molecular weight of H3po4*50)

8 0
3 years ago
Will medal!!!
hjlf
Δ H reaction = q / n where q: amount of heat released and n is number of moles of substance.
q = m . C . ΔT where:
m = mass of substance (g)
C = Specific heat capacity (4.18)
ΔT = change in temperature = 24.25 - 23.16 = 1.09
q = 1000 x 4.18 x 1.09 = 4556 J = 4.556 kJ
number of moles (n) = Molarity (M) x Volume (L)
                                 = 0.185 M x 0.07 L = 0.01295 mole
Δ H = q / n = - (4.556 kJ / 0.01295 mole) = -351.8 kJ / mol
Note: it is exothermic reaction (-ve sign)  i.e. temperature is raised

5 0
3 years ago
Which is true of ultraviolet rays?
Ket [755]
A is true of UV rays.
B is true not of UV rays but rather of visible light.
C is true not of UV rays but rather of microwaves. (unless you actually toast your toast in a toaster like a normal person)
D is true not of UV rays but rather of radio waves.
7 0
3 years ago
Which is of the following is the answer for reactions demonstrates a synthesis reaction?
andrew-mc [135]
The answer is D. This is because a synthesis reaction is when two elements combined to make a product (ex. A+B= AB). Hope this helps!
3 0
3 years ago
What is the percent yield in a reaction between 42.6 g O2 and 49.2 g Al if 72.4 g of Al2O3 is produced?
pashok25 [27]

Answer:

229%

Explanation:

The equation of the reaction is;

4Al(s) + 3O2(g) ----> 2Al2O3(s)

We must first determine the limiting reactant;

Number of moles of Al2O3 produced = mass/molar mass = 72.4g/101.96 g/mol = 0.71 moles

For Al

Number of moles reacted = mass/molar mass = 49.2g/27 g/mol = 1.8 moles

If 4 moles of Al yields 0.71 moles of Al2O3

1.8 moles of Al will yield 1.8 × 0.71/4 = 0.32 moles of Al2O3

For O2

Number of moles reacted = mass/molar mass = 42.6g/32g/mol = 1.33 moles

If 3 moles of O2 yields 0.71 moles of Al2O3

1.33 moles of O2 will yield 1.33 × 0.71/3 = 0.31 moles of Al2O3

Oxygen is the limiting reactant.

% yield = actual yield/ theoretical yield × 100/1

% yield = 0.71 moles/0.31 moles × 100

% yield = 229%

4 0
3 years ago
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