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Kazeer [188]
3 years ago
14

5. an atom that has lost or gained an electron and has become charged

Chemistry
1 answer:
densk [106]3 years ago
5 0

Answer:

It would be compound.

Explanation:

It is this way because if it adds another proton it becomes more positive that nuetral, and if you add an electron it just makes the atom more dense. That is why the answer is compound. Hope this helped :)

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Mohs hardness strontium
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Answer: 5.5

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5 0
3 years ago
What happens when Ethane is treated with bayers reagent​
swat32

Explanation:

यह लेख एक आधार है। जानकारी जोड़कर इसे बढ़ाने में विकिपीडिया की मदद करें।

उस पदार्थ या यौगिक को अभिकर्मक (reagent) कहते हैं जो किसी तंत्र में रासायनिक अभिक्रिया उत्पन्न करने के लिये डाला या मिलाया जाता है। उस पदार्थ को भी अभिकर्मक कहेंगे जिसे यह जांचने के लिये मिलाया जाता है कि कोई अभिक्रिया होती है या नहीं। इस तरह के कुछ वैश्लेषिक अभिकर्मक हैं - फेहलिंग का अभिकर्मक (Fehling's reagent), मिलॉन का अभिकर्मक (Millon's reagent) तथा टॉलीन का अभिकर्मक (Tollens' reagent)।

6 0
3 years ago
A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce
spin [16.1K]

Explanation:

The given data is as follows.

       Volume of lake = 15 \times 10^{6} m^{3} = 15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}

        Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = 5.6 \times 15 \times 10^{9} mg

                                                                    = 84 \times 10^{9} mg

                                                                    = 84 \times 10^{3} kg

Flow rate of river is 50 m^{3} sec^{-1}

Volume of water in 1 day = 50 \times 10^{3} \times 86400 liter

                                          = 432 \times 10^{7} liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are 2.9792 \times 10^{10} mg or 2.9792 \times 10^{4} kg

Flow rate of sewage = 0.7 m^{3} sec^{-1}

Volume of sewage water in 1 day = 6048 \times 10^{4} liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = 1.8144 \times 10^{10} mg or 1.8144 \times 10^{4}kg

Therefore, total concentration of lake after 1 day = \frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l

                                        = 6.8078 mg/l

                 k_{D} = 0.2 per day

       L_{o} = 6.8078

Hence, L_{liquid} = L_{o}(1 - e^{-k_{D}t}

             L_{liquid} = 6.8078 (1 - e^{-0.2 \times 1})  

                             = 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

                                                             = 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0
4 years ago
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