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kenny6666 [7]
3 years ago
13

En una cucharadita de glucosa ( c6 h12 o6) hay en 3'4x10^22 moléculas, ¿Cuántos gramos de glucosa hay?

Chemistry
1 answer:
Alika [10]3 years ago
6 0

Answer:

1 \: mole = 6.02 \times  {10}^{23}  \: molecules \\ ( \frac{3.4 \times  {10}^{22} }{6.02 \times  {10}^{23} } ) \: moles \:  =  \: 3.4 \times  {10}^{22}  \\  = 0.056 \: moles

\small{ \star{ \underline{ \blue{becker}}}}

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For the following aqueous reaction, complete and balance the molecular equation and write a net iconic equatio, making sure to i
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Answer:

Balance molecular equation:

K2CO3(aq) + Sr(NO3)2(aq) → SrCO3(s) + 2KNO3(aq)

Net ionic equation:

CO3∧-2(aq) + Sr∧+2(aq) → SrCO3(s)

Explanation:

Potassium carbonate = K2CO3

Strontium nitrate = Sr(NO3)2

Chemical equation:

K2CO3 + Sr(NO3)2 → SrCO3 + KNO3

Balance chemical equation with physical states:

K2CO3(aq) + Sr(NO3)2(aq) → SrCO3(s) + 2KNO3(aq)

Ionic equation:

2K+(aq) + CO3∧-2(aq) + Sr∧+2(aq) + 2NO∧-3(aq) → SrCO3(s) + 2K+(aq) + 2NO∧-3(aq)

Net ionic equation:

CO3∧-2(aq) + Sr∧+2(aq) → SrCO3(s)

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Spectator ions:

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Consider a 175.7 g sample of the compound manganese(IV) perchlorate.
kramer

Answer:

6.21 moles O

Explanation:

To find the moles of oxygen, you need to (1) convert grams Mn(ClO₄)₄ to moles Mn(ClO₄)₄ (via molar mass) and then (2) convert moles Mn(ClO₄)₄ to moles O (via mole-to-mole ratio from formula subscripts). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units.

Molar Mass (Mn(ClO₄)₄): 452.74 g/mol

1 Mn(ClO₄)₄ = 1 Mn and 4 Cl and 16 O

175.7 g Mn(ClO₄)₄           1 mole                   16 moles O
---------------------------  x  -------------------  x  ---------------------------  =  6.21 moles O
                                       452.74 g            1 mole Mn(ClO₄)₄

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