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2 years ago

How many moles of calcium atoms are in each mole of calcium carbonate?

Chemistry

2 answers:

kaheart [24]2 years ago

7 0

Answer:

You know that one mole of calcium carbonate contains 1 mole of calcium, one mole of carbon, and 3 moles of oxygen. Each mole contains 6.022⋅1023 atoms - this is known as Avogadro's number.

uranmaximum [27]2 years ago

6 0

Answer:

1 mole of calcium, one mole of carbon, and 3 moles of oxygen.

Explanation:

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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

2 years ago

Role and significance of guassion or normal distribution?

kkurt [141]

Normal my guy!!!!!!!!!

5 0

3 years ago

Read 2 more answers

Define unit hi hope u have a great day

Crank

Answer:

Unit is the quantity of a constant magnitude which is used to measure the magnitudes of other quantities of the same nature. 

Explanation:

6 0

2 years ago

Which of the following best explains why the two values in Parts A and B are different? The perceived forces between chlorine ga

BartSMP [9]

Answer:

The high system pressure and relatively large chlorine molecule size.

Explanation:

Having the expression of the ideal gas, and clearing the pressure, we have:

P = nRT/V

Meanwhile, for a non-ideal gas we have the following equation:

P = (nRT / V-nb) - n2a/V2

In this equation, high pressures and low temperatures have an influence on nonideal gases.

Therefore, at high pressures, the molecules in a gas are closer together and have high intermolecular forces. On the other hand, at low temperatures, the kinetic energy of a gas is reduced, so that the intermolecular attractive forces are also reduced.

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2 years ago

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Combustion vapor-air mixtures are flammable over a limited range of concentrations. The minimum volume % of vapor that gives a c

DochEvi [55]

Combustion equation of n-hexane:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles

LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1%

b)
1.1 volume percent required for LFL

1.1% x 1
= 0.0011 m³ of n-hexane required

6 0

2 years ago

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