A flask contains `3.0xx10^(16)` atoms of He. From This `6.6xx10^(15)` atoms of He are removed. The flask now contains atoms of He

Question

A flask contains `3.0xx10^(16)` atoms of He. From This `6.6xx10^(15)` atoms of He are removed. The flask now contains atoms of He

A flask contains `3.0xx10^(16)` atoms of He. From This `6.6xx10^(15)` atoms of He are removed. The flask now contains atoms of He
A. `3.6xx10^(16)`
B. `2.3xx10^(16)`
C. `+3.6xx10^(15)`
D. None of these

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
`3.0xx10^(16)-6.6xx10^(15)`
`=30xx10^(15)-6.6xx10^(15)=(30-6.6)xx10^(15)`
`=23.4xx10^(15)=2.34xx10^(16)=2.3xx10^(16)`
(After rounding off the last digit)

6 views Answered 2023-02-05 15:29