Answer:
Copper(II) chloride (CuCl2) reacts with several metals to produce copper metal or copper(I) chloride (CuCl) with oxidation of the other metal.
Explanation:
Answer:
10.8 days (3 sig.figs.)
Explanation:
All radioactive decay is 1st order decay defined by the expression A = A₀e^-kt
which is solved for time of decay (t) => t = ln(A/A₀) / -k
A = final weight = 1.0 gram
A₀ = initial weight = 16.0 grams
k = rate constant = 0.693/t(1/2) = 0.693/2.69 days = 0.258 days⁻¹
t = ln(1/16) / -0.258da⁻¹ = (-2.77/-0.258) days = 10.74646792 days (calculator)
≅ 10 days (1 sig. fig. based on given 1 gram mass)
To cut this short and for your understanding, ionic bond is formed between metals (mostly right column in periodic table). Covalent bond is formed between non-metals (mostly left column in periodic table). So polar covalent is also a covalent bond but it is polar, which means the shape of molecules are not symmetrical hence maybe an atom in a molecule has most of the electron attracted to it causing itself to be partial negative (since electron are negatively charged) and the other atom has its electron being attracted by others became partial positive. Polar covalent can also be when H atom is binding either to F, O or N (also known as hydrogen bond).
Answer:
4380 mmHg
Explanation:
Boyle's Law can be used to explain the relationship between pressure and volume of an ideal gas. The pressure is inversely related to volume, so if volume decrease the pressure will increase. It can be expressed in the equation as:
P1V1=P2V2
In this question, the first condition is 2L volume and 876 mmHg pressure. Then the system changed into the second condition where the volume is 400ml and the pressure is unknown. The pressure will be:
P1V1= P2V2
876 mmHg * 2L = P2 * 400ml /(1000ml/L)
P2= 876 mmHg * 2L / 0.4L
P2= 4380 mmHg
Answer:
Dolphins use a method called echolocation to detect things such as obstacles and prey in the water. If a dolphin swimming in seawater at 25°C sends a 220-dB click with a frequency of 120.0 Hz, and then detects the reflection of the click exactly one-twentieth of a second later, approximately how far away is the object?
