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2 years ago

Which of the compounds h2c2o4, ca(oh2, koh, and hi, behave as bases when they are dissolved in water?

2 answers:

6 0

H2C2O4, also written as (COOH)2, is Ethandioic acid, and, as evident by the name, is an acid, and not a base
Ca(OH)2 and KOH, Calcium Hydroxide and Potassium Hydroxide, respectively, are bases when dissolved in water, as they are able to release OH- ions.
Hydrogen Iodide, HI, when dissolved in water, is a strong acid. So, that is also not a base.

tatyana61 [14]2 years ago

5 0

Answer: The correct answer is $Ca(OH)_2\text{ and }KOH$

Explanation:

Acids are defined as the chemical species that produces hydrogen ions on dissolving in water.

Bases are defined as the chemical species that produces hydroxide ions on dissolving in water.

We are given:

Chemical compounds having chemical formulas $CH_3COOH,Ca(OH)_2,KOH\text{ and }HI$

$CH_3COOH$ is an organic acid and it produces $CH_3COO^-\text{ and }H^+$ ions

$Ca(OH)_2$ is a base and it produces $Ca^{2+}\text{ and }OH^-$ ions

$KOH$ is a base and it produces $K^{+}\text{ and }OH^-$ ions

HI is an acid because it produces $H^+\text{ and }I^-$ ions

Hence, the correct answer is $Ca(OH)_2\text{ and }KOH$

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Salt water is best described as _______.

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Answer:

Salt water is, A. a mixture

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2 years ago

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William measures a test tube and finds that the mass of the test tube is 5g. He places the lone reactant in the test tube and fi

iogann1982 [59]

86 percent is the percent yield for this experiment if he expected to produce 5g of product.

Explanation:

Given that:

mass of test tube = 5 grams

mass of test tube + reactant is 12.5 grams

mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)

mass of reactant = 12.5 -5

= 7.5 grams

when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.

so mass of product formed = 9.3 - 5

= 4. 3 grams of product is formed (actual yield)

However, he expected the product to be 5 grams (theoretical yield)

Percent yield = $\frac{actual yield}{theoretical yield}$ x 100

putting the values in the formula:

percent yield = $\frac{4.3}{5}$ x 100

= 86 %

86 percent is the percent yield.

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3 years ago

Compound B contains 92.31% carbon and 7.69% hydrogen. Determine the empirical formula of B

tester [92]

Answer:C H

Explanation:

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3 years ago

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A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

Answer: The molecular formula for the compound is $C_6H_6$

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

Step 2: 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 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

Calculating the molar mass of the compound:

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$