Answer:
a. A beta particle has a negative charge. d. A beta particle is a high-energy electron.
Explanation:
Identify the correct descriptions of beta particles.
a. A beta particle has a negative charge. YES. A beta particle is originated in the following nuclear reaction: ¹₀n ⇒ ¹₁H + ⁰₋₁e (beta particle.)
b. A beta particle contains neutrons. NO. It is a electron originated in the nucleus.
c. A beta particle is less massive than a gamma ray. NO. Gamma rays don't have mass while a beta particle has a mass which is half of one thousandth of the mass of a proton.
d. A beta particle is a high-energy electron. YES. Beta particles are nuclear originated hig-energy electrons.
The ideal gas law may be written as
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)
For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol
Therefore
Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm
Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)
First, we must know what happens in the precipitation reaction. This type of reaction is a double replacement reactions. It is consists of two reactant compounds which interchange cations and anions to form two products. One of the products is an insoluble solid called a precipitate. For the precipitation of CaCO₃, there are two consecutive reactions involved:
1. Slaking of quicklime, CaO
CaO + H₂O ⇒ Ca(OH)₂
2. Precipitation
Ca(OH)₂ + CO₂ ⇒ CaCO₃ + H₂O
The ions that make up the H₂O molecule are H⁺ and OH⁻. According to solubility rules, the cation (positively charged ion) is likely to be attracted to an anion (negatively charged ion). Together, they form an ionic bond. This type of bond is when there is a complete transfer of electrons between the two. The Ca²⁺ cation lacks 2 electrons, while the anion OH⁻ has an excess 1 electron. In order to be stable, 1 Ca²⁺ ion and 2 OH⁻ ions must combine.
Therefore, the answer is OH⁻ ion.
The percentage of the sulfur (S) in the compound CuSO₄ is 20.1 %.
<h3>What is the mass percentage?</h3>
The percentage of an element in a compound can be determined as the number of parts by mass of that element present in 100 parts by mass of the given compound.
First, calculate the molecular mass of the given compound by the addition of the atomic masses of all the present elements in the molecular formula. Then, the percentage of the elements can be determined by dividing the total mass of the element by the molar mass of the compound multiplied by 100.
Given, the atomic mass of copper, sulfur, and oxygen is 63.55 g, 32.07 g, and 16.0g respectively.
The molecular mass of CuSO₄ = 63.55 + 32.07 + 4(16.0) = 159.62 g
The mass percentage of the sulphur = (32.07/159.62) × 100 = 20.1 %
Therefore, the mass percentage of the sulfur is equal to 20.1 %.
Learn more about the mass percentage, here:
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The molar concentration of the KI_3 solution is 0.251 mol/L.
<em>Step 1</em>. Write the <em>balanced chemical equation</em>
I_3^(-) + 2S_2O_3^(2-) → 3I^(-) + S_4O_6^(2-)
<em>Step 2</em>. Calculate the <em>moles of S_2O_3^(2-)</em>
Moles of S_2O_3^(2-)
= 27.9 mL S_2O_3^(2-) ×[0.270 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)]
= 7.533 mmol S_2O_3^(2-)
<em>Step 3</em>. Calculate the <em>moles of I_3^(-)
</em>
Moles of I_3^(-) = 7.533 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 3.766 mmol I_3^(-)
<em>Step 4</em>. Calculate the <em>molar concentration of the I_3^(-)
</em>
<em>c</em> = "moles"/"litres" = 3.766 mmol/15.0 mL = 0.251 mol/L