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Flauer [41]
3 years ago
15

Can anyone please help

Chemistry
1 answer:
Naddik [55]3 years ago
7 0
All of them are reactants
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PLEASE HELP <3
frutty [35]
In order for carbon to be stable and have 8 electrons, it must make 4 total covalent bonds.

In prefer for oxygen to be stable and have 8 electrons, it must make 2 covalent bonds.

So, we can deduce that CO2 looks like this:

O=C=O

This molecule has two double bonds.

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8 0
3 years ago
How are ionic bonds formed
Semenov [28]
Ionic bonds are formed when a cation and an anion transfer electrons. The anion gains electrons from the cation to finish its shell, and is usually a nonmetal or a metalloid. A cation gives the anion its electrons to get rid of its partial shell. Cations are metals.
7 0
3 years ago
Read 2 more answers
What kind of solution would you have if it contained 50 grams of sodium chloride in 100 mL of water at 30˚C?
Svetlanka [38]

Answer:

Supersaturated solution.

Explanation:

Hello!

In this case, according to the types of solution in terms of the relative amounts of solute and solvent, we can define a point called solubility at which the amount of solute is no longer dissolved in the solvent; thus, a value of solute/solvent less than the solubility is related to unsaturated solutions, equal to the solubility is related to the saturated solutions and more than the solubility to supersaturated solutions.

Thus, since solubility is temperature-dependent, at 30 °C the solubility of sodium chloride is 36.09 g per 100 mL of water; which means that, since the solution has 50 g of sodium chloride, more than 36.09 g, we infer this is a supersaturated solution.

Best regards!

5 0
3 years ago
A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe
kaheart [24]

Answer:

The specific heat of the alloy C_{a} = 0.37 \frac{KJ}{Kg K}

Explanation:

Mass of an alloy m_{a} = 25 gm

Initial temperature T_{a} = 100°c = 373 K

Mass of water m_{w} = 90 gm

Initial temperature of water T_{w} = 25.32 °c = 298.32 K

Final temperature T_{f} = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

m_{a} C_{a}  [T_{a} - T_{f}] = m_{w} C_w (T_{f} -T_{w} )

25 × C_{a} × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

C_{a} = 0.37 \frac{KJ}{Kg K}

This is the specific heat of the alloy.

6 0
3 years ago
A hot air balloon is filled with 1.31 × 10 6 L of an ideal gas on a cool morning ( 11 ∘ C ) . The air is heated to 121 ∘ C . Wha
victus00 [196]

Answer:

1.82\times 10^6 L is the volume of the air in the balloon after it is heated.

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2} (at constant pressure)

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1= 1.31\times 10^6 L\\T_1=11^oC=(11+273.15)K=284.15K\\V_2=?\\T_2=121^oC=(121+273.15)K=394.15 K

Putting values in above equation, we get:

\frac{1.31\times 10^6 L}{284.15 K}=\frac{V_2}{394.14 K}\\\\V_2=\frac{V_1\times T_2}{T_1}

V_2=1.82\times 10^6 L

1.82\times 10^6 L is the volume of the air in the balloon after it is heated.

4 0
3 years ago
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