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Sedbober [7]
3 years ago
12

you open a can of soda room temperature and here hiss. which of the following factors has changed inside the container

Chemistry
2 answers:
stira [4]3 years ago
8 0
Do u have any options to pick from?
astra-53 [7]3 years ago
4 0

The pressure of gas

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Two glucose molecules can combine to form a disaccharide molecule and what molecule
elena-14-01-66 [18.8K]
And a water molecule, this is called a dehydration synthesis. when 2 molecule combine, a water molecule leave.
5 0
3 years ago
Which of the following statements is TRUE? (A) An ionic bond is much stronger than most covalent bonds.(B) Once dissolved in wat
Ainat [17]

Answer:

A)  An ionic bond is much stronger than most covalent bonds.

Explanation:

D) Ionic compounds have high melting points causing them to be solid at room temperature, and conduct electricity when dissolved in water. Covalent compounds have low melting points and many are liquids or gases at room temperature.

C) An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a cation, which is usually a metal, and an anion, which is usually a nonmetal. A covalent bond involves a pair of electrons being shared between atoms.

A) Covalent bonds are stronger if you compare with ionic molecules, because their molecular orbital overlap is bigger. However, ionic molecules form lattices, thus the energy to break this lattice bond is stronger hence the ionic bond is stronger.

4 0
4 years ago
A 0.0100 M solution of AgNO3 has a concentration of Ag ion equivalent to
qaws [65]

Answer:

0.0100M of AgNO3 contains 0.0100M of Ag+

Explanation:

AgNO3 when ionized yields Ag+ and NO3-. This means that the amount of AgNO3 in solution is equivalent to the amount of Ag+ and NO3- in that same solution.

1M of AgNO3 solution produces 1M of Ag+  

1M of AgNO3 solution produces 1M of NO3-

This occurs because of the complete ionization of AgNO3 in solution, allowing complete dissolution of the compound.

6 0
3 years ago
An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the
Brrunno [24]

Answer:

Approximately 81.84\%.

Explanation:

Balanced equation for this reaction:

{\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm  NaCl}\, (aq) + {\rm CaCO_{3}}\, (s).

Look up the relative atomic mass of elements in the limiting reactant, \rm CaCl_{2}, as well as those in the product of interest, \rm CaCO_{3}:

  • \rm Ca: 40.078.
  • \rm Cl: 35.45.
  • \rm C: 12.011.
  • \rm O: 15.999.

Calculate the formula mass for both the limiting reactant and the product of interest:

\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}.

\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the quantity of the limiting reactant (\rm CaCl_{2}) available to this reaction:

\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}.

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant (\rm CaCl_{2}) and the product ({\rm CaCO_{3}}) are both 1. Thus:

\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1.

In other words, for every 1\; \rm mol of \rm CaCl_{2} formula units that are consumed, 1\; \rm mol\! of \rm CaCO_{3} formula units would (in theory) be produced. Thus, calculate the theoretical yield of \rm CaCO_{3}\! in this experiment:

\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}.

Calculate the theoretical yield of this experiment in terms of the mass of \rm CaCO_{3} expected to be produced:

\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}.

Given that the actual yield in this question (in terms of the mass of \rm CaCO_{3}) is 13.19\; \rm g, calculate the percentage yield of this experiment:

\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}.

6 0
3 years ago
What are atoms? Types of atoms?
shtirl [24]

Answer:

The definition of atom: The Atom is the basic unit of matter.

Types of atoms: There are 3 types of atoms, protons, neutrons, and electrons.

6 0
3 years ago
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