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pickupchik [31]
2 years ago
12

What is the overall balanced equation for the precipitation reaction occurring between silver nitrate and calcium bromide? hints

what is the overall balanced equation for the precipitation reaction occurring between silver nitrate and calcium bromide? 2agno3(aq)+cabr2(aq)?2agbr(s)+ca(no3)2(aq) agno3(aq)+cabr2(aq)?agbr(s)+ca(no3)2(aq) 2agno3(aq)+cabr2(aq)?2agbr(aq)+ca(no3)2(aq) 2agno3+cabr2?2agbr+ca(no3)2?
Chemistry
1 answer:
aleksklad [387]2 years ago
4 0
The answer:

<span>the overall balanced equation for the precipitation reaction occurring between silver nitrate and calcium bromide:
</span>CaBr2 + AgNO3-----------> AgBr2 + CaNO3, so among the given choices, the
true answe is 
<span>agno3(aq)+cabr2(aq) (as reactants)</span>
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One crystalline form of silica (SiO2) has a cubic unit cell, and from X-ray diffraction data it is known that the cell edge leng
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Answer:

8 Silicon atom are present in unit cell.

16 oxygen atoms are present unit cell.

Explanation:

Number of atoms in unit cell = Z =?

Density of silica = tex]2.32 g/cm^3[/tex]

Edge length of cubic unit cell = a  = 0.700 nm = 0.700\times 10^{-7} cm

1 nm=10^{-7} cm

Molar mass of Silica  = 28.09 g/mol+16.00\times 2=60.09 g/mol

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

2.32 g/cm3=\frac{Z\times 60.09 g/mol}{6.022\times 10^{23} mol^{-1}\times (0.700\times 10^{-7}cm)^{3}}

Z = 8

1 silicon is 2 oxygen atoms. then 8 silicon atoms will be 16 oxygen atoms.

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2 years ago
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During which time interval does the substance exist as both a liquid and a solid
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14. What is the pH of a 0.24 M solution of sodium propionate, NaC3H502, at 25°C? (For
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Answer:

9.1

Explanation:

Step 1: Calculate the basic dissociation constant of propionate ion (Kb)

Sodium propionate is a strong electrolyte that dissociates according to the following equation.

NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻

Propionate is the conjugate base of propionic acid according to the following equation.

C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻

We can calculate Kb for propionate using the following expression.

Ka × Kb = Kw

Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰

Step 2: Calculate the concentration of OH⁻

The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.

[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M

Step 3: Calculate the concentration of H⁺

We will use the following expression.

Kw = [H⁺] × [OH⁻]

[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M

Step 4: Calculate the pH of the solution

We will use the definition of pH.

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5 0
2 years ago
Marianne designs an experiment involving electrically charged objects. She wants to know which objects will be attracted to a ne
svet-max [94.6K]

Answer:

When one object is rubbed against another, static electricity can be created. This is because the rubbing creates a negative charge that is carried by electrons. The electrons can build up to produce static electricity. For example, when you shuffle your feet across a carpet, you are creating many surface contacts between your feet and the carpet, allowing electrons to transfer to you, thereby building up a static charge on your skin. When you touch another person or an object, you can suddenly discharge the static as an electrical shock.

Similarly, when you rub a balloon on your head it causes opposite static charges to build up both on your hair and the balloon. Consequently, when you pull the balloon slowly away from your head, you can see these two opposite static charges attracting one another and making your hair stand up.

Materials

• Balloon

• An object made out of wool (such as a sweater, scarf, blanket or ball of yarn)

• Stopwatch

• A wall

• A partner (optional)

Preparation

• Blow up the balloon and tie off the end.

• Have your partner prepare to use the stopwatch.

Procedure

• Hold the balloon in a way that your hand covers as little of its surface area as possible, such as by using only your thumb and pointer finger or by gripping the balloon by its neck where it is tied off.

• Rub the balloon on the woolly object once, in one direction.

• Hold the balloon up on the wall with the side that was rubbed against the wool facing the wall, then release it. Does the balloon stay stuck on the wall? If the balloon stays stuck, have your partner immediately start the stopwatch to time how long the balloon remains bound to the wall. If the balloon does not stick, move to the next step.

• Touch the balloon to a metal object. Why do you think this is important to do?

• Repeat the above process but each time increase the number of times you rub the balloon on the woolly object. Rub the balloon in the same direction each time. (Do not rub the balloon back and forth.)

Observations and results

In general, did the balloon stick to the wall for a longer amount of time as you increased the number of times you rubbed the balloon on the woolly object?

Wool is a conductive material, which means it readily gives away its electrons. Consequently, when you rub a balloon on wool, this causes the electrons to move from the wool to the balloon's surface. The rubbed part of the balloon now has a negative charge. Objects made of rubber, such as the balloon, are electrical insulators, meaning that they resist electric charges flowing through them. This is why only part of the balloon may have a negative charge (where the wool rubbed it) and the rest may remain neutral.

When the balloon has been rubbed enough times to gain a sufficient negative charge, it will be attracted to the wall. Although the wall should normally have a neutral charge, the charges within it can rearrange so that a positively charged area attracts the negatively charged balloon. Because the wall is also an electrical insulator, the charge is not immediately discharged. However, because metal is an electrical conductor, when you rub the balloon against metal the extra electrons in the balloon quickly leave the balloon and move into the metal so the balloon is no longer attracted and does not adhere.

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST

6 0
2 years ago
Read 2 more answers
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