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Shkiper50 [21]
2 years ago
12

All simple metal with acid reactions are ____ displacement reactions.

Chemistry
1 answer:
erastova [34]2 years ago
6 0

Answer:

single displacement

Explanation:

the metal replaces the hydrogen of the acid to for a salt and hydrogen gas

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Given that a for HBrO is 2. 8×10^−9 at 25°C. What is the value of b for BrO− at 25°C?
lara [203]

If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

<h3>What is base dissociation constant? </h3>

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 2.8× 10^(-9)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{2.8×10^(-9) }

= 3.5× 10^(-6)

Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

DISCLAIMER: The above question have mistake. The correct question is given as

Question:

Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?

learn more about base dissociation constant:

brainly.com/question/9234362

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7 0
2 years ago
What is the percent yield of O2 if 10.2 g of O2 is produced from the decomposition of 17.0 g of H2O?
yawa3891 [41]

The balanced chemical reaction will be:

2H2O = 2H2 + O2

<span>We are given the amount of water used in the decomposition reaction. This will be our starting point.</span>

<span>17.0 g H2O</span> (1 mol  H2O/ 18.02 g H2O) (1 mol O2/2 mol <span>H2O</span>) ( 32.00 g O2/1mol O2) = 15.09 g O2

Percent yield = actual yield / theoretical yield x 100

<span>Percent yield =10.2 g / 15.09  g x 100</span>

Percent yield = 67.58%

5 0
3 years ago
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How can you tell if glucose is present?
rosijanka [135]

Answer:

Plants use photosynthesis to make glucose. Glucose is also know as sugar. You can tell it is present if the plant receives sunlight as well as water.

Explanation:

MARK BRAINLIEST

4 0
3 years ago
Are humans pro-karyotic or eukaryotic?
barxatty [35]

Answer:

eukaryotic

Explanation:

all human cells—including those found in the brain, the heart, the muscles, and so on—are also eukaryotic.

3 0
3 years ago
What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2
atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}

Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

Mole of AlCl_3 = Mole of Al * Mole ratio

Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = 133.34 \frac{g}{mol}

Mass of AlCl3 can be calculated using mole formula as:

1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}

Multiplying both side by 133.34 \frac{g}{mol}

1.32 mole  * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34  \frac{g}{mol}

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

Mole of AlCl_3 = mole of Cl_2 * mole ratio

Mole of AlCl_3 = 0.55  mole  * \frac{2}{3}

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}

Multiplying both side by

133.34 \frac{g}{mol}

0.367 mol of AlCl_3 * 133.34 \frac{g}{mol}  = \frac{mass(g)}{133.34\frac{g}{mol}}   * 133.34 \frac{g}{mol}

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0
3 years ago
Read 2 more answers
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