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3 years ago

What is the density of a block of gold that occupies 1000 ml and has a mass of 3.5 kg? Show your work

Chemistry

1 answer:

tiny-mole [99]3 years ago

5 0

Answer:

Density of block of gold is <em>3.5 g/cm³.</em>

Explanation:

Given data:

Volume of block = 1000 mL

Mass of block = 3.5 kg (3.5×1000 = 3500 g)

Density of block = ?

Solution:

Density of substance is calculated by dividing the mass of substance over its volume.

Formula:

d = m/v

d = 3500 g/ 1000 mL

d = 3.5 g/mL

d = 3.5 g/cm³ (1ml = 1cm³)

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This material has a density of 1 g/cm3.
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Which condition increases the number of collisions between reactant molecules in a given volume?

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Answer:

Increasing Surface Area

Explanation:

A greater surface area (meaning more, smaller particles) allows for more opportunity for particles to collide. On the other hand, decreasing temperature and removing a catalyst would only decrease the number of collisions, and the clumping option doesn't make much sense. Hope this helps!

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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:

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

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3 years ago

What happens if the metal you throw in is MORE REACTIVE than the<br> metal ion in solution?

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A displacement reaction Would occur in this situation

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According to the kinetic molecular theory particles of matter

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Answer:

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Explanation:

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