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4 years ago

Separete the ALKALI from the following bases :

Chemistry

1 answer:

V125BC [204]4 years ago

3 0

Answer:

Ammonium hydroxide, NH₄OH

Magnesium hydroxide, Mg(OH)₂

Sodium hydroxide, NaOH

Lithium hydroxide, LiOH

Explanation:

A base is a substance which neutralizes acids to produce salt and water. Bases are hydroxide or oxides of metals. Bases may be soluble or insoluble in water. Bases generally have a bitter taste and turn red litmus paper or indicator red.

Alkalis are bases which are soluble in water. They form the hydroxide of the alkali metals or alkaline earth metals in solution and they ionize to produce hydroxide ions. They are slippery to touch and turn red litmus blue being bases.

Therefore, all alkalis are bases but not all bases are alkalis. Insoluble bases are not alkalis.

From the given chemical compounds the alkalis present in the list are:

Ammonium hydroxide, NH₄OH; since it is soluble in water and produces hydroxide ions

Magnesium hydroxide, Mg(OH)₂; since it is slightly soluble in water and produces hydroxide ions

Sodium hydroxide, NaOH; since it is soluble in water and produces hydroxide ions

Lithium hydroxide, LiOH; since it is soluble in water and produces hydroxide ions

CuO(copper oxide) is a base but not an alkali as it does not produce hydroxide ions.

Zn(OH)2 (zinc hydroxide) is amphoteric and is insoluble

MgO(magnesium oxide) is a base but not an alkali as it does not produce hydroxide ions.

Na2O(sodium oxide) is a base but not an alkali as it does not produce hydroxide ions.

CoO(cobalt oxide) is a base but not an alkali as it does not produce hydroxide ions.

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How many moles of methane occupy a volume of 2.00 l at 50.0°c and 0.500 atm answer?

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Data Given:
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Volume = V = 2.0 L

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Solution:
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P V = n R T
Solving for n,
n = P V / R T

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3 years ago

Calculate: (a) the weight (in lbf) of a 30.0 lbm object. (b) the mass in kg of an object that weighs 44N. (c) the weight in dyne

belka [17]

Answer:

a) 965,1 lbf

b) 4,5 kg

c) 1,33 * 10^6 dynes

Explanation:

Mass of an object refers to the amount of mattter it cotains, it can be expressed it gr, kg, lbm, ton, etc.

Weight of an object refers to a force, and is the measurement of the pull of gravitiy on an object. It may be definide as the mass times the acceleration of gravity.

w=mg

In Planet Earth, the nominal "average" value for gravity is 9,8 m/s² (in the International System) or 32,17 ft/s² (in the FPS system).

To solve this problem we'll use the following conversion factors:

1 lbf = 1 lbm*ft/s²

1 N = 1 kg*m/s²

1 dyne = 1 gr*cm/s² and 1 N =10^5 dynes

1 ton = 907,18 kg

1 k = 1000 gr

a) m = 30 lbm

$w = 30 lbm * 32,17 ft/s^{2} = 965, 1 \frac{lmb*ft}{s^{2} } = 965,1 lbf$

b) w = 44 N

First, we clear m of the weight equation and then we replace our data.

$m = \frac{w}{g} = \frac{44 N}{9,8 \frac{m}{{s}^{2}} } = 4,5 kg$

c) m = 15 ton

$m=15 ton * \frac{907,18 kg}{1 ton} = 13607,7 kg \\ w = mg = 13607,7 kg * 9,8 m/s2 = 133355,5 N * \frac{10^{5} dynes }{1 N} = 1,33 * 10^{6}dynes$

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4 years ago

Consider the neutralization reaction 2 HNO 3 ( aq ) + Ba ( OH ) 2 ( aq ) ⟶ 2 H 2 O ( l ) + Ba ( NO 3 ) 2 ( aq ) A 0.120 L sample

k0ka [10]

Answer:

The concentration of the HNO3 solution is 0.150 M

Explanation:

Step 1: Data given

Volume of the unknown HNO3 sample = 0.120 L

Volume of the 0.200 M Ba(OH)2 = 45.1 mL

Step 2: The balanced equation

2HNO3 + Ba(OH)2 ⟶ Ba(NO3)2 + 2H2O

Step 3: Calculate moles Ba(OH)2

moles Ba(OH)2 = molarity * volume

moles Ba(OH)2 = 0.200 M * 0.0451 L

moles Ba(OH)2 = 0.00902 moles

Step 4: Calculate moles of HNO3

For 1 mole of Ba(OH)2 we need 2 moles of HNO3

For 0.00902 moles of Ba(OH)2 we need 2*0.00902 = 0.01804 moles

Step 5: Calculate molarity of HNO3

molarity = moles / volume

molarity = 0.01804 / 0.120 L

Molarity = 0.150 M HNO3

The concentration of the HNO3 solution is 0.150 M

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