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Lemur [1.5K]
3 years ago
8

The term “electricity” in referring to static electricity, means what

Chemistry
1 answer:
sladkih [1.3K]3 years ago
5 0
Is an imbalance of electric charges within or on the surface of a material. The charge remains until it is able to move away by means of an electric current or electrical discharge. Static electricity is named in contrast with current electricity, which flows through wires or other conductors and transmits energy.[1]
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The word computer dates to the 1600s, so if you want to know the trems etymology, you will have to start your research in the time period.

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Classify each item by matching as organic
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3 0
3 years ago
. If the half-life of cesium-133 is 30 years, how much of a 600g sample of cesium-133 will be left after
e-lub [12.9K]

Answer:

Caesium (55Cs) has 40 known isotopes, making it, along with barium and mercury, one of the elements with the most isotopes. The atomic masses of these isotopes range from 112 to 151. Only one isotope, 133Cs, is stable. The longest-lived radioisotopes are 135Cs with a half-life of 2.3 million years, ... It constitutes most of the radioactivity still left from the Chernobyl accident ...

3 0
3 years ago
A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a
Semenov [28]

<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol

  • The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = \frac{1}{2}\times 0.01670=0.00835mol of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

  • Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

  • The chemical equation for the reaction of metal (forming M^{3+} ion) and sulfuric acid follows:

2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = \frac{2}{3}\times 0.0556=0.0371mol of metal

  • To calculate the molar mass of metal for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0
3 years ago
You have 350 mL of 3.4 M hydrochloric acid (HCl). How many grams of HCl gas are dissolved? Bonus: what is the volume of the HCl
Crank

Answer:

1. 43.44g of HCl

2. 26.67 L of HCl

Explanation:

1) Molarity of a solution = number of moles (n) ÷ Volume (V)

According to the provided information in this question,

V = 350 mL = 350/1000 = 0.350L

Molarity = 3.4 M

Using Molarity = n/V

3.4 = n/0.350

n = 3.4 × 0.350

n = 1.19mol

Using the formula below to calculate the mass of HCl;

mole = mass/molar mass

Molar mass of HCl = 1 + 35.5 = 36.5g/mol

mole = mass/MM

mass = 1.19 mol × 36.5g/mol

mass = 43.44g of HCl

2) At STP, HCl has a pressure of 1atm, a temperature of 273K

V = ?

n = 1.19 mol

R = 0.0821 Latm/molK

Using PV = nRT

V = nRT/P

V = 1.19 × 0.0821 × 273/1

Volume = 26.67L

5 0
3 years ago
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