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A flask contains 2.0xx10^(13) molecules ...

A flask contains `2.0xx10^(13)` molecules of `CO_(2)`. To this `1.5xx10^(14)` molecules of `CO_(2)` are added.
The number of molecules in the flask now is

A

`3.5xx10^(13)`

B

`3.5xx10^(14)`

C

`1.7xx10^(14)`

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To find the total number of molecules of CO₂ in the flask after adding more molecules, we can follow these steps: ### Step 1: Identify the initial number of molecules The initial number of CO₂ molecules in the flask is given as: \[ 2.0 \times 10^{13} \text{ molecules} \] ### Step 2: Identify the number of molecules added The number of CO₂ molecules added to the flask is: \[ 1.5 \times 10^{14} \text{ molecules} \] ### Step 3: Convert both numbers to the same power of ten To add the two quantities, we need to express them with the same exponent. The smaller exponent is \(10^{13}\). We can rewrite \(1.5 \times 10^{14}\) as: \[ 1.5 \times 10^{14} = 15.0 \times 10^{13} \] ### Step 4: Add the two quantities Now we can add the two quantities: \[ 2.0 \times 10^{13} + 15.0 \times 10^{13} = (2.0 + 15.0) \times 10^{13} = 17.0 \times 10^{13} \] ### Step 5: Convert the result to standard scientific notation The result \(17.0 \times 10^{13}\) can be expressed in standard scientific notation as: \[ 1.7 \times 10^{14} \] ### Final Answer Thus, the total number of CO₂ molecules in the flask now is: \[ \boxed{1.7 \times 10^{14}} \] ---

To find the total number of molecules of CO₂ in the flask after adding more molecules, we can follow these steps: ### Step 1: Identify the initial number of molecules The initial number of CO₂ molecules in the flask is given as: \[ 2.0 \times 10^{13} \text{ molecules} \] ...
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Knowledge Check

  • The weight of 2.01xx10^(23) molecules of CO is-

    A
    `9.3 g`
    B
    `7.2g`
    C
    `1.2g`
    D
    `3g`
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    A
    760.0 mm of Hg
    B
    76.0 mm of Hg
    C
    7.6 mm of Hg
    D
    0.76 mm of Hg
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    A
    `3.6xx10^(16)`
    B
    `2.3xx10^(16)`
    C
    `+3.6xx10^(15)`
    D
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