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3 years ago

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Dichloromethane contains strong carbon - hydrogen and carbon- chlorine bonds. Despite the presence of these strong bonds. dichlo

romethane is a liquid...explain why??
can some one help me ??
50 points

1 answer:

const2013 [10]3 years ago

4 0

Answer:

vot. tofu p f focpxe9d3dotc

Explanation:

circeicirct pupdtoxkr

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Selenium (Se) has 30 protons and 20 neutrons in each atom, therefore its atomic

allochka39001 [22]

Answer:

30, 50

Explanation:

Hello,

In this since an element's atomic number is equal to the number of protons in its atom, we can infer that selenium's atomic number is 30. Moreover, due to the fact the the neutrons are equal to the atomic mass minus the atomic number or the number of protons, by knowing the number of neutrons we compute the atomic as follows:

$neutrons=mass-protons\\\\mass=neutrons+protons\\\\mass=30+20\\\\mass=50a.m.u$

Thus, answer is 30, 50.

Best regards.

8 0

3 years ago

How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

8 0

3 years ago

The rate of decomposition of acetaldehyde, CH_3 CHO(g), into CH_4(g) and CO(g) in the presence of I_2(g) at 800 K follows the ra

FrozenT [24]

Answer:

Explanation:

Catalyst is I2 . Because I2 is reacted with starting material in step 1 and generated in second step

Rate limiting step is step 1. Because in rate equation CH3CHO and I2 is mentioned. Hence the overall rate of reaction is depending CH3CHO and I2. Rate limiting step is step 1

5 0

3 years ago

A student has the following data: V1 = 822 mL; T1 = 75.0 °C; T2 = 25.0 °C. She uses this data to calculate V1 and gets -274 mL.

luda_lava [24]

Answer to this question is C. Regarding the volume.

4 0

3 years ago

If the chemist mistakenly makes 250 mL of solution instead of the 200 mL, what molar concentration of sodium nitrate will the ch

Evgesh-ka [11]

Answer:

0.120M is the concentration of the solution

Explanation:

<em>Assuming the mass of sodium nitrate dissolved was 2.552g</em>

<em />

Molar concentration is an unit of concentration widely used in chemsitry defined as the moles of solute (In this case NaNO3) in 1L of solution.

To find this question we must find the moles of NaNO3 in 2.552g. With this mass and the volume (250mL = 0.250L) we can find molar concentration as follows:

<em>Moles NaNO3 -Molar mass: 84.99g/mol-</em>

2.552g * (1mol / 84.99g) = 0.0300 moles NaNO3

<em>Molar concentration:</em>

0.0300 moles NaNO3 / 0.250L =

<h3>0.120M is the concentration of the solution</h3>

7 0

3 years ago