Which two structures will provide a positive identification of a plant cell under a microscope? A.) Lysosomes, cell wall. B.) large central vacuole, cell wall. C.) large central vacuole ribosomes. D.)nucleoid, chloroplasts.

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Answer 1

Answer: The Right Answer Is D.) Nucleoid chloroplasts.

Answer 2

Answer: the answer is B large central vacuole and cell wall. they are the easiest/biggest things to see under a microscope to identify a plant cell

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Which particles are equal in number for an atom with a neutral charge?

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electrons and protons

The value of the Solubility Product Constant for lead phosphate is ____________Write the reaction that corresponds to this Ksp value.

_______(Aq,S,L) +_______(Aq,S,L) <-------->_______(Aq,S,L) +_______(Aq,S,L)

Ksp values are found by clicking on the "Tables" link.

Use the pull-down menus to specify the state of each reactant or product.

If a box is not needed leave it blank.

Answers

Answer: The reaction for the $K_(sp)$ value of lead phosphate is given below and the value of solubility product for the same is $3.0\rightarrow 10^(-44)$

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio. It is expressed as $K_(sp)$

The chemical formula of lead phosphate is $Pb_3(PO_4)_2$

The equation for the hydration of the lead phosphate is given as:

$Pb_3(PO_4)_2(s)+H_2O(l)\rightarrow 3Pb^(2+)(aq.)+2PO_4^(3-)(aq.)$

The solubility product of lead phosphate is $3.0\rightarrow 10^(-44)$ . This means that it is highly insoluble in water as the solubility product is very very low.

Hence, the reaction for the $K_(sp)$ value of lead phosphate is given above and the value of solubility product for the same is $3.0\rightarrow 10^(-44)$

Consider the following reaction: Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)If you react an excess of Pb(NO3)2with 26.3 g of NaCl, and you isolate 52.1 g of PbCl2, what is your percent yield?

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Answer:

$\large \boxed{84.7 \, \%}$

Explanation:

Mᵣ: 58.44 278.11

Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃

m/g: 26.3

1. Moles of NaCl

$\text{Moles of NaCl} = \text{26.3 g NaCl} * \frac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} = \text{0.4505 mol NaCl}$

(b) Moles of PbCl₂

$\text{Moles of PbCl${_2}$} = \text{0.4505 mol NaCl} * \frac{\text{1 mol PbCl${_2}$}}{\text{2 mol NaCl}} = \text{0.2253 mol PbCl${_2}$}$

(c) Theoretical yield of PbCl₂

$\text{Mass of PbCl${_2}$} = \text{0.2253 mol PbCl${_2}$} * \frac{\text{278.11 g PbCl${_2}$}}{\text{1 mol PbCl${_2}$}} = \text{61.52 g PbCl${_2}$}$

(d) Percent yield

$\text{Percent yield} = \frac{\text{ actual yield}}{\text{ theoretical yield}} * 100 \,\% = \frac{\text{52.1 g}}{\text{61.52 g}} * 100 \, \% = \mathbf{84.7 \,\%}\n\n\text{The percent yield is $\large \boxed{\mathbf{84.7 \, \% }}$}$

What is the solubility in pure water of ba(io3)2 in moles per liter at 25 ˚c? [ksp (25 ˚c) = 6.0  10–10]?

Answers

Ba(IO₃)₂(s) partially dissociates in water as Ba²⁺(aq) and IO₃⁻(aq).
Ba(IO₃)₂(s) ⇄ Ba²⁺(aq) + 2IO₃⁻(aq)
Initial
Change -X +X 2X
Equilibrium X 2X

Ksp = [Ba²⁺(aq)] x [IO₃⁻(aq)]²
6.0 x 10⁻¹⁰ = X * (2X)²
6.0 x 10⁻¹⁰ = 4X³
X = 5.313 x 10⁻⁴ mol/L

Hence, the solubility of the Ba(IO₃)₂(s) is 5.313 x 10⁻⁴ mol/L

What reaction conditions most effectively conver a cabocxylic acid to a methly ester?

Answers

Answer:

Esterification reaction

Explanation:

When we have to go from an acid to an ester we can use the esterification reaction. On this reaction, an alcohol reacts with a carboxylic acid on acid medium to produce an ester and water. (See figure).

In this case, we need the methyl ester, therefore we have to choose the appropriate alcohol, so we have to use the methanol as reactive if we have to produce the methyl ester.

Saved Propane burns in air according to the equation C3Ha(g 502lg)3CO2) + 4H20(g) What volume of O2 in liters would be required if 15.0 L of propane burns, assuming that all of the gases are under the same conditions? Short Answer Toolbar navigation E I E B IUS EA This question will be sent to your Instructor for grading. 20 of 25 l Next > Prev nere to search

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Answer: 75 liters of $O_2$ in liters would be required if 15.0 L of propane burns, assuming that all of the gases are under the same conditions.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 Lat STP and contains avogadro's number $6.023* 10^(23)$ of particles.

To calculate the number of moles, we use the equation:

$\text{Number of moles of propane}=\frac{\text{Given volume}}{\text{Molar volume}}=(15.0L)/(22.4L)=0.67moles$

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

According to stoichiometry:

1 mole of propane combines with = 5 moles of oxygen

Thus 0.67 moles of propane combine with = $(5)/(1)* 0.67=3.35moles$

Volume of $O_2=moles* {\text {Molar Volume}}=3.35* 22.4L=75L$

Thus 75 liters of $O_2$ in liters would be required if 15.0 L of propane burns, assuming that all of the gases are under the same conditions.

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