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Petunia needs 66.48 grams of potassium hydroxide (KOH) to run a reaction in class. How many moles of potassium hydroxide does sh

Kaylis [27]

Answer:

She needs 1.19 moles of potassium hydroxide.

Explanation:

Being the molar mass of the elements:

K= 39 g/mole
O= 16 g/mole
H= 1 g/mole

then the molar mass of potassium hydroxide is:

KOH= 39 g/mole + 16 g/mole + 1 g/mole= 56 g/mole

Being the mass of one mole of a substance, which can be an element or a compound, you can apply the following rule of three: if 56 g of KOH are present in 1 mole, 66.48 g of KOH in how many moles of the compound are they?

$moles of KOH=\frac{66.48 grams*1 mole}{56 grams}$

moles of KOH= 1.19

She needs 1.19 moles of potassium hydroxide.

4 0

3 years ago

For each reaction, calculate how many moles of the barium product you will produce using stoichiometry and the balanced reaction

ankoles [38]

Answer:

Explanation:

given that

mass of Ba(NO3)2 = 1.40g

mass of NH2SO3H = 2.50 g

1)to determine the mole of Ba(NO3)2

2) to determine the mass of all three product formed in the reaction

reaction

Ba(NO3)2 + 2NH2SO3H → Ba(NH2SO3)2 + 2HNO3

Solution

we calculate the molar mass of each species by using their atomic masses

BA = 137.33g/mol

N = 14g/mol

O= 16g/mol

H = 1g/mol

S = 32g/mol

calculation

Ba(NO3)2 = Ba + 2N + 6O

= 137.33 + 2X 14 + 6 X 16

= 261.33g/mol

NH2SO3H = N + 3H + S+ 3O

=14 + 3X1 + 32 + 3X 16

= 97g/mol

Ba(NH2SO3)2 = Ba + 2N + 4H +2S +6O

= 137.33 + 2 X 14 + 4 X1 + 2X32 + 6 X 16

= 329.33g/mol

HNO3 = H + n + 3O

= 1 + 14 + 3 X 16

= 63g/mol

3 0

4 years ago

What are properties that metals all have? I need 3

krok68 [10]

Answer:

solid at room temperature (with the exception of mercury)

usually shiny.

high melting point.

good conductor of heat.

good conductor of electricity.

low ionization energies.

low electronegativities.

malleable – able to be pounded into sheets.

Explanation:

8 0

3 years ago

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The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T

motikmotik

Answer :

(a) The rate law for the reaction is:

$\text{Rate}=k[OCl^-]^1[I^-]^1$

(b) The value of rate constant is, $60.4M^{-1}s^{-1}$

(c) rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$OCl^-+I^-\rightarrow OI^-+Cl^-$

Rate law expression for the reaction:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

where,

a = order with respect to $OCl^-$

b = order with respect to $I^-$

Expression for rate law for first observation:

$1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(1)

Expression for rate law for second observation:

$2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(2)

Expression for rate law for third observation:

$2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1$

Dividing 1 from 3, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1$

Thus, the rate law becomes:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

a = 1 and b = 1

$\text{Rate}=k[OCl^-]^1[I^-]^1$

Now, calculating the value of 'k' (rate constant) by using any expression.

$1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})$

$k=60.4M^{-1}s^{-1}$

Now we have to calculate the rate for a reaction when concentration of $OCl^-$ and $I^-$ is $1.8\times 10^{-3}M$ and $6.0\times 10^{-4}M$ respectively.

$\text{Rate}=k[OCl^-][I^-]$

$\text{Rate}=(60.4M^{-1}s^{-1})\times (1.8\times 10^{-3}M)(6.0\times 10^{-4}M)$

$\text{Rate}=6.52\times 10^{-5}Ms^{-1}$

Therefore, the rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

8 0

3 years ago

Bases can ________. release only hydrogen ions take up only hydrogen ions release only hydroxide ions both take up hydrogen ions

Lapatulllka [165]

Bases can both take up hydrogen ions and release hydroxide ions. This differs from an acid, which releases a hydrogen ion. Combining acids and bases can produce a lot of heat during the reaction. If the acid and base are of the same strength, the mixture will be neutralized.

4 0

4 years ago

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