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3 years ago

8

The density of a substance is 4.8 g/mL. What is the volume of the sample that is 19.2 g?

1 answer:

mixas84 [53]3 years ago

5 0

Answer:

The answer is 4

Explanation:

You divide the grams by the volume which will give you the density

so 19.2 divided by 4.8 equals 4

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In physical science applying a force over a displacement is called _______.

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In physical science applying a force over a displacement is called work.

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4 years ago

A sample of oxygen gas occupies 3.60 liters at a pressure of 1.00 atm. If temperature is held constant, what will be the volume

evablogger [386]

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4 years ago

Read 2 more answers

At 25 °C, what is the hydroxide ion concentration, [OH-], in an aqueous solution with a hydrogen ion concentration of

Otrada [13]

The hydroxide ion concentration, [OH-] = 2.439 x 10⁻¹⁰

<h3>Further explanation</h3>

Equilibrium of water and the ions :

$\tt 2H_2O(l)\rightarrow H_3O^+(aq)+OH^-(aq)$

$\tt Kc=\dfrac{[H_3O^+][OH^-]}{H_2O}$

Kc[H₂O] is the ion equilibrium constant for water ⇒ Kw.

The product of the concentrations of these ions (H₃O⁺ and OH⁻)at equilibrium at 25°C is 1.0 x 10⁻¹⁴ M

[H₃O⁺] [OH⁻] = 1.0 x 10⁻¹⁴ M = Kw

The hydrogen ion concentration of [H⁺]=4.1 x 10⁻⁵ M, so the hydroxide ion concentration, [OH-] :

$\tt 4.1\times 10^{-5}\times [OH^-]=1.10^{-14}\\\\(OH^-]=\dfrac{1.10^{-14}}{4.1\times 10^{-5}}\\\\(OH^-]=2.439\times 10^{-10}$

6 0

3 years ago

In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) + H2O (g) ⇌

Oksana_A [137]

Answer: Equilibrium constant is 0.70.

Explanation:

Initial moles of $CO$ = 0.35 mole

Volume of container = 1 L

Initial concentration of $CO=\frac{moles}{volume}=\frac{0.35moles}{1L}=0.35M$

Initial moles of $H_2O$ = 0.40 mole

Volume of container = 1 L

Initial concentration of $H_2O=\frac{moles}{volume}=\frac{0.40moles}{1L}=0.40M$

equilibrium concentration of $CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M$ [/tex]

The given balanced equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initial conc. 0.35 M 0.40M 0 0

At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}$

$K_c=\frac{x\times x}{(0.40-x)(0.35-x)}$

we are given : (0.35-x)= 0.18

x = 0.17

Now put all the given values in this expression, we get :

$K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}$

$K_c=0.70$

Thus the value of the equilibrium constant is 0.70.

5 0

3 years ago