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3 years ago

A sample of marble has a volume of 6 cm3 and a density of 2.76 g/cm3. What is its mass?

Chemistry

2 answers:

zhuklara [117]3 years ago

6 0

Answer:

16.56g

Explanation:

6*2.76=16.56...

Eduardwww [97]3 years ago

3 0

Answer:

16.56 grams

Explanation:

The density can be found using the following formula:

$d=\frac{m}{v}$

If we rearrange the formula for m, the mass, we should multiply both sides by v.

$d*v=\frac{m}{v} *v$

$d*v=m$

The mass can be found by multiplying the density and the volume. The density is 2.76 grams per cubic centimeter. The volume is 6 cubic centimeters.

$d= 2.76 g/c^3\\v=6 cm^3$

Substitute the values into the formula.

$m= d*v$

$m= 2.76 g/cm^3 * 6 cm^3$

Multiply. When you multiply the 2 "cm^3" will cancel each other out.

$m= 2.76 g* 6$

$m= 16.56 g$

The mass of the marble is 16.56 grams.

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zavuch27 [327]

Answer:

$K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}$

Explanation:

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In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

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The suitable equilibrium constant turns out:

$K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}$

Or in terms of the initial equilibrium constant:

$K_2=K_1^2$

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Best regards.

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3 years ago

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Alecsey [184]

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8 0

3 years ago

What kind of questions CANNOT be answered by chemistry?

Sergeu [11.5K]

Answer:

d. why matter exists

Explanation:

The kind of questions that chemistry CANNOT answer is "why matter exists".

In Chemistry, question of how the properties, composition and structure of substances are is answered. Also, the transformations that these substances undergo, and the energy that they release or absorbe during the transformation processes are revealed in chemistry.

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3 years ago

Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e

igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

Explanation:

1) Moles of NaCl , $n_1=1 mol$

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Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

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$p=17.19 Torr$

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose , $n_1=1mol$

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$p'=17.19 Torr$

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

3 0

3 years ago

What is the mass, in grams, of a sample of 1.20 × 1022 atoms of mercury (Hg)? Show your work or explain the steps that you used

o-na [289]

Atomic mass Hg = 200.59 u.m.a

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mass Hg = ( 1.20x10²² ) x 200.59 / 6.02x10²³

mass Hg = 2.407x10²⁴ / 6.02x10²³

= 3.998 g of Hg

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3 years ago

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