How many Neutrons are in a ^15N isotope?

A bottle of wine contains 0.21 moles of ethanol,C2H5OH. The volume of the solution is 0.1 L. Calculate the molarity.

Yuliya22 [10]

Answer:

2.1mol/L

Explanation:

Number of moles = 0.21 moles

Volume = 0.1L

Molarity of a substance is the number of moles of solute dissolved in a volume of solvent (L)

Molarity = number of moles / volume of solvent

Molarity = 0.21 / 0.1

Molarity = 2.1mol/L

8 0

3 years ago

I NEED THE ANSWER IMMEDIATELY!!! Pls helppp

Wewaii [24]

Answer:

turgor pressure can be done in a lab or a self test.

turgor pressure is key to the plant’s vital processes. It makes the plant cell stiff and rigid. Without it, the plant cell becomes flaccid. Prolonged flaccidity could lead to the wilting of plants.

Turgor pressure is also important in stomate formation. The turgid guard cells create an opening for gas exchange. Carbon dioxide could enter and be used for photosynthesis. Other functions are apical growth, nastic movement, and seed dispersal.

Explanation:

salt is bad for turgor pressure.

Turgidity helps the plant to stay upright. If the cell loses turgor pressure, the cell becomes flaccid resulting in the wilting of the plant.
The wilted plant on the left has lost its turgor as opposed to the plant on the right that has turgid cells.

7 0

3 years ago

Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r

Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺ + 2e⁻ → Mg

6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺ + 2e⁻ → Mg

Si → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺ + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺ + 4e⁻ → 2Mg

6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺ + 4e⁻ + 6OH⁻ + Si → 2Mg + SiO₃²⁻ + 4e⁻ + 3 H₂O

7 0

3 years ago

In redox reactions, _____________ occurs when electrons are lost by a molecule. ______________ occurs when electrons are gained

madam [21]

Answer:

Oxidation, Reduction

Explanation:

A redox reaction is a short form for reduction-oxidation.

Reduction is a term which means that electron is gained while oxidation is a term which means that electron Is lost.

The species that gain electron is known as the oxidizing agent while the species losing electrons is known as the reducing agent

6 0

3 years ago

Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin

STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of $KI$ and $AgNO_3$ .

$\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}$

Molar mass of KI = 166 g/mole

$\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole$

and,

$\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole$

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

$KI+AgNO_3\rightarrow KNO_3+AgI$

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of $AgNO_3$

So, 0.0178 mole of KI react with 0.0178 mole of $AgNO_3$

From this we conclude that, $AgNO_3$ is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgI$