All categories

3 years ago

What is an acid?????????????

Chemistry

2 answers:

zimovet [89]3 years ago

7 0

Answer:

An acid is- A chemical substance that neutralizes alkalis, dissolves some metals, and turns litmus red. Or, typically, a corrosive or sour-tasting liquid of this kind.

Explanation:

katrin [286]3 years ago

5 0

Answer:

a chemical substance that neutralizes alkalis, dissolves some metals, and turns litmus red; typically, a corrosive or sour-tasting liquid of this kind.

Explanation:

You might be interested in

Which human activity would most likely decrease the amount of carbon in the atmosphere?

Allushta [10]

The removal of trees would most likely decrease the amount of carbon in the atmosphere.

Please note that it is useful to add the options provided with the question, in order to get a most accurate answer and have your question answered quicker.

Hope this helps!!

6 0

3 years ago

Read 2 more answers

Compare the reactivity between lithium and sodium towards oxygen.

Scrat [10]

Reactivity of Group 1 and 2 elements increases as you go down the periodic table. So sodium is more reactive than lithium. Sodium will react with oxygen forming Na2O (sodium oxide). Lithium forms lithium oxide (Li2O).

7 0

2 years ago

Define alkanes and give examples

LenKa [72]

<span>the simplest organic molecules, consisting of only carbon and hydrogen and with only single bonds between carbon atoms.
Ex. Methane, Ethan, Propane, Butane
</span>

3 0

2 years ago

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

3 years ago

2. A student has a centrifuge tube containing 14.0 g of t-butanol and is asked to make a 1.2 m solution of ethanol/t-butanol. Ho

Vaselesa [24]

Answer:

0.774g of ethanol

0.970mL of ethanol

Explanation:

Molality is an unit of concentration defined as the ratio between moles of solute and kg of solvent.

In the problem, you need to prepare a 1.2m solution of ethanol (Solute) in t-butanol (solvent).

14.0g of butanol are <em>0.014kg </em>and as you want to prepare the 1.2m solution, you need to add:

0.014kg × (1.2moles / kg) = 0.0168 moles of solute = Moles of ethanol

To convert moles of ethanol to mass you require molar mass (Molar mass ethanol, C₂H₅OH = 46.07g/mol). Thus, mass of 0.0168 moles are:

0.0168moles Ethanol ₓ (46.07g / mol) =

<h3>0.774g of ethanol</h3>

And to convert mass in g to mL you require density of the substance (Density of ethanol = 0.798g/mL):

0.774g ₓ (1mL / 0.798g) =

<h3>0.970mL of ehtanol</h3>

8 0

2 years ago