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Volgvan
3 years ago
14

How many lithium atoms are in 4li2o?

Chemistry
2 answers:
choli [55]3 years ago
6 0

Answerthere are 8

Explanation:

erica [24]3 years ago
3 0

Answer:

8

Explanation:

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Cold packs are used to treat minor injuries. Some cold packs contain NH4NO3(s) and a small packet of water at room temperature b
Anastasy [175]
These types of reactions are called endothermic reactions. This is when the system absorbs the heat from the surrounding. In this case, the system is the chemical reaction and the environment is the direct vicinity of the chemical reaction. You notice that the pack becomes cold because heat from the surrounding is being absorb. 

The bonds involved in this substance includes ionic bonding and polar covalent. 
6 0
4 years ago
A solution has a volume of 500 ml and contains 0.570 mol of nacl what is the molarity
vichka [17]

Answer: 1.14 M

Explanation:

500 mL is equal to 0.5 L.

So, the molarity is (0.570)/(0.5) = 1.14 M

7 0
2 years ago
HELP ASAP!!!
lisabon 2012 [21]
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5 0
3 years ago
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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
A gas cylinder contains exactly 1 mole of oxygen gas (O2). How many molecules of oxygen are in the cylinder?.
Svet_ta [14]

Answer:

4. 01 × 1022 molecules 6. 02 × 1023 molecules 9. 03 × 1024 molecules 2. 89 × 1026 molecules.

Explanation:A gas cylinder contains exactly 1 mole of oxygen gas (O2). How many molecules of oxygen are in the cylinder?

4 0
2 years ago
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