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3 years ago

Which of the following increases the concentration of OH- ions?

Chemistry

2 answers:

ludmilkaskok [199]3 years ago

7 0

Answer:

Basee is the answer hope its write xd

Trava [24]3 years ago

3 0

Answer:

B: Base

Explanation:

An acid will increase the concentration of hydrogen (H+) ions in a solution, and a base would increase the concentration of hydroxide (OH–) ions in a solution. So, Base is the answer. :)

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Pls help, How is stoichiometry used to calculate amounts of substances in a chemical reaction?

Elden [556K]

Answer:

Thus, to calculate the stoichiometry by mass, the number of molecules required for each reactant is expressed in moles and multiplied by the molar mass of each to give the mass of each reactant per mole of reaction. The mass ratios can be calculated by dividing each by the total in the whole reaction.

Explanation: Stoichiometry is the field of chemistry that is concerned with the relative quantities of reactants and products in chemical reactions. For any balanced chemical reaction, whole numbers (coefficients) are used to show the quantities (generally in moles ) of both the reactants and products.

7 0

2 years ago

Use the balanced equation below to answer the question that follows

natta225 [31]

It’s c hdjdkdmnfvdheidk

4 0

2 years ago

. Consider the following half-reactions:

Vedmedyk [2.9K]

d. Fe(s) and Al(s)

<h3>Further explanation</h3>

In the redox reaction, it is also known

Reducing agents are substances that experience oxidation

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction

The electrodes which are easier to reduce than hydrogen (H), have E cells = +

The electrodes which are easier to oxidize than hydrogen have a sign E cell = -

So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)

The metal : d. Fe(s) and Al(s)

7 0

3 years ago

Glass is an example of a(n) ________ solid

romanna [79]

Amorphous is the answer. A glass is an amorphous solid.

3 0

3 years ago

Read 2 more answers

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

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 $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

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 $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago