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3 years ago

Is gastric (stomach) fluid an acid or a base?

1 answer:

3 0

Answer: the such thing that we call gastric acid, is made/produced by the cells that srebwithi any lining of our stomac, they are coupled in places like feedback system that extend to the acid production when it is needed.

other cells that are within our stomach will bicarbonat, at the base to buffer the fluid making sure that it doesn’t become too acidic

so yes it is

Explanation: hope this helped plz mark brainest

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Classify each of these soluble solutes as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. Solutes formula hydroch

matrenka [14]

Compounds which on dissolving in water gets completely dissociates into its ions are known as strong electrolytes whereas compounds which on dissolving in water gets partially dissociates into its ions are known as weak electrolytes.

Substances which gives solution on dissolving in water and do not dissociates into ions also does not conduct electric current are known as nonelectrolyte.

Hydrochloric acid, $HCl$

On adding $HCl$ (strong acid) in water, it will completely dissociates into ions ( $H^{+}$ and $Cl^{-}$ ) and thus, it is a strong electrolyte.

Sodium hydroxide, $NaOH$

On adding $NaOH$ (strong base) in water, it will completely dissociates into ions ( $Na^{+}$ and $OH^{-}$ ) and thus, it is a strong electrolyte.

Formic acid, $HCOOH$

On adding $HCOOH$ (weak acid) in water, it will partially dissociates into ions ( $H^{+}$ and $HCOO^{-}$ ) and thus, it is a weak electrolyte.

Methyl amine, $CH_3NH_2$

On adding $CH_3NH_2$ (weak base) in water, it will partially dissociates into ions ( $CH_3NH_3^{+}$ and $OH^{-}$ ) and thus, it is a weak electrolyte.

Potassium chloride, $KCl$

On adding $KCl$ in water, it will completely dissociates into ions ( $K^{+}$ and $Cl^{-}$ ) and thus, it is a strong electrolyte.

Ethanol, $C_2H_5OH$

On adding $C_2H_5OH$ in water, it will not dissociates into ions and thus, it is a nonelectrolyte.

Sucrose, $C_{12}H_{22}O_{11}$

On adding $C_{12}H_{22}O_{11}$ in water, it will not dissociates into ions and thus, it is a nonelectrolyte.

3 0

3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

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 $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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.

For calculating the mass of hydrogen:

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:

Step 1: 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 = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

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 0.00775 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 = $\frac{0.00775}{0.00775}=1$

Step 3: 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 $C_6H_4O_2$

6 0

3 years ago

An aqueous basic solution has a concentration of 0. 050 m and kb is 4. 4 × 10^-4. What is the concentration of hydronium ion in

Alex_Xolod [135]

An aqueous basic solution has a concentration of 0. 050 m and kb is 4. 4 × 10^-4, the concentration of hydronium ion in this solution (m) is 2.234 × 10⁻¹² M.

Methylamine is an amine which is an organic weak base. Its chemical formula is CH₃NH₂. When it undergoes hydrolysis wherein water is acting as an acid, the reaction would be

CH₃NH₂ + H₂O ⇆ CH₃NH₃ + OH⁻

Then, we use the ICE analysis which stands for Initial-Change-Equilibrium.

CH₃NH₂ + H₂O ⇆ CH₃NH₃ + OH⁻

Initial 0.05 - 0 0

Change -x +x +x

----------------------------------------------------------------------------

Equilibrium 0.05-x x x

Then, we use the equation for the equilibrium constant of basicity.

Kb = [CH₃NH₃][OH⁻]/[CH₃NH₂] = 4.4×10⁻⁴

4.4×10⁻⁴ = [x][x]/[0.05-x]

[x] = 4.4756×10⁻³

Here, variable x denotes the number of moles of the substance which is involved in the reaction. Since the equilibrium amount of OH⁻ is equal to x, then the concentration of OH⁻ is also 4.4756×10⁻³. Thus,

pOH = -log[OH⁻]

pOH = -log[4.4756×10⁻³] = 2.35

The relationship between pOH and pH is that pH + pOH = 14. Thus,

pH = 14 - 2.35 = 11.65

pH = -log[H⁺]

11.65 = -log[H⁺]

[H⁺] = 2.234 × 10⁻¹² M

Thus, we find the concentration of solution is 2.234 × 10⁻¹² M.

Learn more about aqueous solution: brainly.com/question/11097800

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5 0

2 years ago

Which two atoms could be the "before'' and ''after'' of an alpha ejection

spin [16.1K]

The answer would be uranium and thorium. When an alpha ejects a particle, it will create a new atom. So, when uranium ejects an alpha particle, it will produce thorium. They call this process as the alpha decay. Alpha decay often happens on atoms that are abundant nuclei such as uranium, radium, and thorium.

4 0

3 years ago

Atoms with a low electronegativity,like lithium,have a weak attractive force for electrons because

Murrr4er [49]

Answer:

they're close to filling their outer shell, fulfilling the octet rule

Explanation:

3 0

3 years ago