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solong [7]
3 years ago
12

Classify these salts as acidic, basic, or neutral. KCL, NH4Br, K2CO3, NaCN, LiClO4? This is what I put but got it wrong: KCl = b

asic NH4Br = acidic K2CO3 = basic NaCN = basic LiClO = Neutral Can someone me what I did wrong, thanks.
Chemistry
2 answers:
Alenkinab [10]3 years ago
8 0

KCl : neutral

NH₄Br : acidic

K₂CO₃ : basic

NaCN : basic

LiClO : basic

<h3>Further explanation</h3>

Salt is an ion compound consisting of metal cations and residual acid anions. Cation salts can come from a base and anions come from acids. So that each salt has a base component and an acid component

So there are 4 properties of salt

1. salts from strong acids and strong bases are neutral

2. salts from strong acids and weak bases are acidic

3. salts from weak acids and strong bases are basic

4. salts of weak acids and weak bases depend on the value of the ionization constant of the acid and its base (Ka and Kb)

Ka> Kb is acidic

Ka <Kb is basic

Ka = Kb is neutral

  •  1. KCL,

salt of strong acid HCl and strong base of KOH, neutral

  • 2. NH₄Br,

salt of strong acid HBr and weak base NH₄OH, acidic

  •  3. K₂CO₃,

salt of H₂CO₃ weak acid and strong base KOH, basic

  • 4. NaCN,

salt of a weak acid HCN and a strong base NaOH, basic

  • 5. LiClO

salt of a weak acid HClO and a strong base LiOH, basic

<h3>Learn more </h3>

the Bronsted-Lowry definition of an acid

brainly.com/question/3201146

the Bronsted-Lowry definition of a base

brainly.com/question/4083753

the acids and the bases in the chemical equation according to Bronsted-Lowry Theory

brainly.com/question/12308171

Keywords:  acid, base, salt

MaRussiya [10]3 years ago
3 0
In classifying salts as neutral, acidic, or basic, it is important to take note of the strength of the acids and bases that they come from. A strong acid and strong base produce a neutral salt. A weak acid and strong base produce a basic salt. A strong acid and weak base produce an acidic salt. So the answers must be:

KCl = neutral (from HCl and KOH)
NH4Br = acidic (from NH4 and HBr)
K2CO3 = basic (from KOH and H2CO3)
NaCN = basic (from NaOH and HCN)
LiClO = basic (from LiOH and HClO)
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<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_2H_2O_4 and C_6H_4O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=1.03g

Mass of H_2O=0.14g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, \frac{12}{44}\times 1.03=0.28g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, \frac{2}{18}\times 0.14=0.016g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = \frac{0.023}{0.00775}=2.96\approx 3

For Hydrogen  = \frac{0.016}{0.00775}=2.06\approx 2

For Oxygen  = \frac{0.00775}{0.00775}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is C_3H_{2}O_1=C_3H_2O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

n=\frac{108.10g/mol}{54g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2

Thus, the empirical and molecular formula for the given organic compound is C_3H_2O and C_6H_4O_2

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An important reaction that takes place in a blast furnace during the production of iron is the formation of iron metal and CO2 f
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Answer: Mass of Fe_2O_3 required to form 930 kg of iron is 1328 kg

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

For iron:

Given mass of iron = 930 kg = 930000 g  (1kg=1000g)

Molar mass of iron = 56 g/mol

Putting values in equation 1, we get:

\text{Moles of iron}=\frac{930000g}{56g/mol}=16607mol

The chemical equation for the  production of iron  follows:

Fe_2O_3+3CO\rightarrow 2Fe+3CO_2

By Stoichiometry of the reaction:

2 moles of iron are  produced by =  1 mole of Fe_2O_3

So, 16607 moles of iron will be produced by = \frac{1}{2}\times 16607=8303moles of Fe_2O_3

Now, calculating the mass of Fe_2O_3 from equation 1, we get:

Mass of Fe_2O_3 = moles\times {\text {molar mass}}=8303\times 160=1328480g=1328kg

Thus mass of Fe_2O_3 required to form 930 kg of iron is 1328 kg

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How many electrons does phosphorous (P) need to gain to have a stable outer electron level?
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Suppose that 0.50 grams of barium-131 are administered orally to a patient. Approximately how many milligrams of the barium woul
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Two months later 13.8 milligrams of the barium-131 still be radioactive.

<h3>How is the decay rate of a radioactive substance expressed ? </h3>

It is expressed as:

A = A_{0} \times (\frac{1}{2})^{t/T}

where,

A = Amount remaining

A₀ = Initial Amount

t = time

T = Half life

Here

A₀ = 0.50g

t  = 2 months = 60 days

T = 11.6 days  

Now put the values in above expression we get

A = A_{0} \times (\frac{1}{2})^{t/T}

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   = 0.50 \times (\frac{1}{2})^{5.17}

   = 0.50 × 0.0277

   = 0.0138 g

   = 13.8 mg          [1 mg = 1000 g]

Thus from the above conclusion we can say that Two months later 13.8 milligrams of the barium-131 still be radioactive.

Learn more about the Radioactive here: brainly.com/question/2320811

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Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Suppose that 0.50 grams of ban that 0.50 grams of barium-131 are administered orally to a patient. Approximately many milligrams of the barium would still be radioactive two months later? The half-life of barium-131 is 11.6 days.

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