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dexar [7]
3 years ago
11

An acorn falls from a tree and accelerates at 9.8 m/s2. It hits the ground in 0.67 s. What is the speed of the acorn when it hit

s the ground?
Chemistry
1 answer:
ivolga24 [154]3 years ago
8 0
Acceleration means that while it's falling, its speed will increase by 9.8 m/s
every second.  That's the acceleration of everything that falls near the Earth's
surface, no matter how heavy or light the object is.

In 0.67 seconds, then, the speed of the acorn increases from zero to

                 (0.67 x 9.8) = <u>6.57 m/s</u> .
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PLEASE HELP!<br><br> what are the steps that occur during the combustion of hydrogen?
Softa [21]

Answer:

Explanation:

Combustion releases energy in a single step in the form of light and heat. Whereas in respiration, energy is released in steps and is stored in the form of ATP.

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3 years ago
PLEASEEEE HELP MEEEEE
fiasKO [112]

Answer:

3.82 x 10²¹ molecules As₂O₃

Explanation:

To find the amount of molecules arsenic (III) oxide (As₂O₃), you need to (1) convert kg to lbs, then (2) convert g As₂O₃ to moles As₂O₃ (via molar mass), and then (3) convert moles to molecules (via Avogadro's number).

1 kilogram = 2.2 lb

Molar Mass (As₂O₃): 2(74.992 g/mol) + 3(15.998 g/mol)

Molar Mass (As₂O₃): 197.978 g/mol

Avogadro's Number:

6.022 x 10²³ molecules = 1 mole

0.0146 g As₂O₃            1 kg                  189 lb
------------------------  x  ---------------  x  ------------------  x   ................
         1 kg                     2.2 lb          

       1 mole                6.022 x 10²³ molecules
x  ------------------  x  ---------------------------------------  = 3.82 x 10²¹ molecules As₂O₃
      197.978 g                        1 mole

6 0
2 years ago
Read 2 more answers
Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(
kkurt [141]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) 2Sr(s)+O_2(g)\rightarrow 2SrO(s)    \Delta H_1=-1184kJ

(2) SrO(s)+CO_2(g)\rightarrow SrCO_3(s)     \Delta H_2=-234kJ      ( × 2)

(3) CO_2(g)\rightarrow C(s)+O_2(g)     \Delta H_3=394kJ    ( × 2)

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ

Hence, the \Delta H^o_{rxn} for the reaction is 72 kJ.

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4 0
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