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4 years ago

Which of the following describes a base? Select all that apply.

Chemistry

2 answers:

Elanso [62]4 years ago

7 0

I believe your answers are B, tastes bitter, and D, slippery feel. An example would be a bar of soap.

Ne4ueva [31]4 years ago

4 0

<h2>Answer </h2>

Option B, C, D - Tastes Bitter, Highly Reactive, Slippery Feel

Explanation

A base is described by a bitter taste, reactive and feels slippery. Option A is not correct this is because the base obtain curtain specific features. Following are the characteristics of a base is:

It possesses a bitter taste.

It is slippery.

Many bases react with acids and by reacting with acids it forms salts.

Bases turn red litmus paper blue.

Bases possess metal oxides or hydroxides.

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Based on the nebular theory of the solar system formation, explain how our solar system is formed.

Alika [10]

the nebular theory states that the solar systems developing rotations revolve around sun. it is located in the milky-way. revolving around the sun, with gravity pulling it, an interstellar cloud of dust and gas, is formed. (i.e. the nebula that we are speaking)

this formation is a protostar. the idea in the theory states that the nebula has flattened into a disk called the "protoplanetary disk"; and over time, the debreis eventually formed this disk. (i.e. creating the milky-way)

hopefully this helped, thank you!! :D

6 0

3 years ago

Please help ASAP! I’m lost right now.

SCORPION-xisa [38]

1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1

Chromium is strange because it moves on to the 4s orbital instead of filling the 3d orbital with that last electron. Tricky.

Mark as brainliest if this helped! :)

3 0

3 years ago

What is the molality of a solution of water and KCl if the freezing point of the solution is –3°C? (Kf = 1.86°C/m; molar mass of

Elina [12.6K]

Some of the solutions exhibit colligative properties. These properties depend on the amount of solute dissolved in a solvent. These properties include freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering. Calculations are as follows:

ΔT(freezing point) = (Kf)mi
3 = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg

4 0

3 years ago

Read 2 more answers

To answer this question, you will need to write the balanced equation and set up a BCA table. Using appropriate rounding rules,

Elenna [48]

Water decomposes when electrolyzed to produce hydrogen and oxygen gas. If 2.5 grams of water were decomposed 1.04 grams of oxygen will be formed.

BCA table:

2 $H_{2}$ O ⇒ $H_{2}$ + $O_{2}$

B 0.13 0 + 0

C -0.13 0.065 + 0.065

A 0 0.065

Explanation:

Balanced equation for water decomposition into hydrogen and oxygen gases

2 $H_{2}$ O ⇒ $H_{2}$ + $O_{2}$

B 0.13 0 + 0

C -0.13 0.065 + 0.065

A 0 0.065

Number of moles of water = $\frac{mass}{atomic mass of 1 mole}$

mass = 2.5 grams

atomic mass= 18 grams

number of moles can be known by putting the values in the formula,

n = $\frac{2.5}{18}$

= 0.13 moles

2 moles of water gives one mole of oxygen on decomposition

so, 0.13 moles of water will give x moles of oxygen on decompsition

$\frac{1}{2}$ = $\frac{x}{0.13}$

x = 0.065 moles of oxygen will be formed.

moles to gram will be calculated as

mass =number of moles x atomic mass

= 0.065 x 16

= 1.04 grams of oxygen.

7 0

4 years ago

Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2

IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at $75^{o}F$ and 14.6 psia.

$\varepsilon$ = 0.00015 ft, Flow rate, (Q) = 48000 $ft^{3}/m$

(a) Formula to calculate hydraulic radius $(r_{H})$ is as follows.

$r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

= $\frac{2 \times 1}{2(1) + 2(2)}$

= $\frac{1}{3}$ ft

Formula for equivalent diameter is as follows.

$D_{eq} = 4 \times r_{H}$

= $4 \times \frac{1}{3} ft$

= $\frac{4}{3}$ ft

(b) Formula for velocity floe is as follows.

Q = VA

V = $\frac{Q}{A}$

= $\frac{48000}{2 \times 1}$ ft/min

= 24000 ft/min

(c) Formula to calculate Reynold's number is as follows.

$R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

= $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$ (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

= 3224559.6

(d) Formula to calculate pressure drop $(\Delta P)$ is as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

= $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

= 6.238 $lb/ft^{2}$

5 0

3 years ago