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3 years ago

Why chloride described as an extended structure?

Chemistry

1 answer:

Amanda [17]3 years ago

6 0

Answer:

Explanation:

Examples of extended structures could include sodium chloride or diamonds. Examples of molecular-level models could include drawings, 3D ball and stick structures, or computer representations showing different molecules with different types of atoms.]

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for the reaction HCI+NaOH➡️NaCi+H2O, how many moles of hydrochloric acid are required to produce 150 g. of water?

Ne4ueva [31]

Answer: 8.33 mol of HCl (Hydrochloric Acid)

Explanation:

150 g H2O x __1 mol__ x __1 mol HCl__ = 8.33 mol of HCl

18.016 g 1 mol H2O

6 0

3 years ago

How many grams of Pb are contained in a mixture of 0.135 kg each of PbCl(OH) and Pb2Cl2CO3?

Dahasolnce [82]

Hey there!:

Given the mass of PbCl(OH) :

0.135 Kg = 0.135 Kg*(1000g / 1Kg) = 135 g

Molecular mass of PbCl(OH) = 207+35.5+16+1 = 259.5 g / mol

Atomic mass of Pb = 207 g/mol

Hence mass of Pb in 135 g PbCl(OH) :

(207 g Pb / 259.5 g PbClOH) * 135g PbClOH =

0.79768 * 135 => 107.68 g of Pb

For Pb2Cl2CO3 :

Given the mass of Pb2Cl2CO3 :

0.135 Kg = 0.135 Kgx(1000g / 1Kg) = 135 g

Molecular mass of Pb2Cl2CO3 = 2*207+2*35.5+12+3*16 = 545 g / mol

Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3 = 2*207 = 414 g

Hence mass of Pb in 135 g Pb2Cl2CO3:

(414 g Pb / 545 g PbClOH) * 135g PbClOH =

0.75963 * 135 => 102.55 g of Pb2Cl2CO3

Hope that helps!

8 0

3 years ago

Read 2 more answers

What is the molarity of a NaOH solution if 28.2 mL of a 0.355 M H2SO4 solution is required to neutralize a 25.0-mL sample of the

Verdich [7]

Answer:

[NaOH} = 0.4 M

Explanation:

In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.

(H₂SO₄, is considered strong, but the first deprotonation is weak)

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.

In the equivalence point we know mmoles of base = mmoles of acid

Let's finish the excersise with the formula

25 mL . M NaOH = 28.2 mL . 0.355M

M NaOH = (28.2 mL . 0.355M) / 25 mL → 0.400

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

Your favorite beach has 42 large sand dunes throughout the course of the year wind erosion destroys eight sand dunes and create

a_sh-v [17]

42- 8 = 34 sand dunes must be the answer

hope so it helps

8 0

3 years ago