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sladkih [1.3K]
3 years ago
12

A negatively charged balloon has 2.6 µc of charge. how many excess electrons are on this balloon? the elemental charge is 1.6 ×

10−19
c. answer in units of electrons.
Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
5 0
The given 2.6 µC of charge is due to a buildup of electrons, each of which has a charge of 1.6 x 10^-19 C. The 2.6 <span>µC is equivalent to 2.6 x 10^-6 C, so we can divide this by the individual charge of an electron:
</span>2.6 x 10^-6 C / 1.6 x 10^-19 (C/electron) = 1.625 x 10^13 electrons
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An atom that gains or loses an electron has a net electric charge and is called a/n
Fofino [41]
That would be an ion, so A.
8 0
2 years ago
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How many mL of a stock 50% (w/v) KNO3 solution are needed to prepare 250 mL of a 20% (w/v) KNO3 solution?
Andre45 [30]

Answer:

100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.

Explanation:

In the given question it is mentioned that

     S1=50%

      V2=250ml

      S2= 20%

We all know that

                     V1S1=V2S2

                     ∴V1=  V2×S2÷S1

                     ∴V1=  V2S2×1/S1

                      ∴V1= 250×20÷50

                       ∴V1= 100ml

 

6 0
3 years ago
What happens to the temperature and density of the material between points B and C?
makvit [3.9K]

Answer:

Temperature decreases and density increases

Explanation:

Let us remember that density of a material increases as the temperature of the material decreases. So the cooler a material becomes, the denser it becomes also.

Between points B and C, the material rapidly cools down and the temperature decreases accordingly. This ultimately results in an increase in density since cooler materials are denser than hot materials.

7 0
3 years ago
Silver nitrate and calcium chloride react to form calcium nitrate and silver chloride. Write the equation.
8090 [49]

Answer:

Explanation:

2AgNO3 + CaCl2 ----> 2AgCl +  Ca(NO3)2

6 0
3 years ago
A solution is prepared by mixing 2.17 g of an unknown non-electrolyte with 225.0 g of chloroform. The freezing point of the resu
Deffense [45]

Answer:

The molar mass of the unknown non-electrolyte is 64.3 g/mol

Explanation:

Step 1: Data given

Mass of an unknown non-electrolyte = 2.17 grams

Mass of chloroform = 225.0 grams

The freezing point of the resulting solution is –64.2 °C

The freezing point of pure chloroform is – 63.5°C

kf = 4.68°C/m

Step 2: Calculate molality

ΔT = i*kf*m

⇒ ΔT = The freezing point depression = T (pure solvent) − T(solution) = -63.5°C + 64.2 °C = 0.7 °C

⇒i = the van't Hoff factor = non-electrolyte = 1

⇒ kf = the freezing point depression constant = 4.68 °C/m

⇒ m = molality = moles unknown non-electrolyte / mass chloroform

0.7 °C = 1 * 4.68 °C/m * m

m = 0.150 molal

Step 3: Calculate moles unknown non-electrolyte

molality = moles unknown non-electrolyte / mass chloroform

Moles unknown non-electrolyte = 0.150 molal * 0.225 kg

Moles unknown non-electrolyte = 0.03375 moles

Step 4: Calculate molecular mass unknown non-electrolyte

Molar mass = mass / moles

Molar mass = 2.17 grams / 0.03375 moles

Molar mass = 64.3 g/mol

The molar mass of the unknown non-electrolyte is 64.3 g/mol

6 0
3 years ago
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