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butalik [34]
4 years ago
12

0.517 = what’s this in scientific notation

Chemistry
2 answers:
JulijaS [17]4 years ago
4 0
The answer is 5.17 x 10^-1
Stells [14]4 years ago
4 0

this is the answre belwo

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Morgarella [4.7K]
B) chemical bond i believe is the correct answer
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Period 4 of this periodic table is highlighted. What happens as you move from right to left across the period?
ivanzaharov [21]
The elements change from metals to nonmetals.
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3 years ago
Read 2 more answers
When 0.491 grams of a protein were dissolved in 44 mL of benzene at 24.4 degrees C, the osmotic pressure was found to be 50.9 to
Ira Lisetskai [31]

Answer:

4057.85 g/mol

Explanation:

Hello, the numerical procedure is shown in the attached file.

- In this case, since we don't have the density of the protein, we must assume that the volume of the solution is solely given by the benzene's volume, in order to obtain the moles of the solute (protein).

-Van't Hoff factor is assumed to be one.

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4 0
3 years ago
The solubility of water in diethyl ether has been reported to be 1.468 % by mass.' Assuming that 23.0 mL of diethyl ether were a
melomori [17]

Explanation:

As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.

Hence, amount of diethyl ether present will be calculated as follows.

                          (100ml - 1.468 ml)

                        = 98.532 ml

So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.

Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.

          Amount of water = \frac{1.468 ml \times 23 ml}{98.532 ml}

                                       = 0.3427 ml

Now, when magnesium dissolves in water then the reaction will be as follows.

                Mg + H_{2}O \rightarrow Mg(OH)_{2}

Molar mass of Mg = 24.305 g

Molar mass of H_{2}O = 18 g

Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.

           Amount of Mg = \frac{24.305 g \times 0.3427 ml}{18 g}  

                                    = 0.462 g

   

6 0
4 years ago
1. If I have 45 L of He in a balloon at 25 degrees celsius and increase the temperature of the
Greeley [361]

Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).

We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.

Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.  

(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).

7 0
3 years ago
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