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IrinaK [193]
3 years ago
5

Which element has similar properties to lithium? Explain your reasoning:

Chemistry
1 answer:
Sphinxa [80]3 years ago
4 0

Answer:

sodium/potassium/rubidium/caesium/francium

Explanation:

all are group I elements, so they all have similar properties

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Thorium is an element that occurs naturally in the earth's crust. The only naturally occurring isotope of thorium is 232Th and i
Murrr4er [49]

A, radium-228. Alpha radiation is essentially just two protons and two neutrons leaving the atom, causing the atomic number to drop by two (since the protons left) and the mass number to drop by four (since two neutrons and protons left).

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3 years ago
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The two south pole ends of two magnets are touching. Which of the following can be concluded?
larisa86 [58]

C.Work was required by an outside force.

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Where are non metals found in the periodic table
Romashka [77]

The metals are to the left of the line (except for hydrogen, which is a nonmetal), the nonmetals are to the right of the line, and the elements immediately adjacent to the line are the metalloids.

hope it helps...!!!

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2 years ago
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A teacher puts two tablets in a glass of water. The teacher
ehidna [41]

This question is incomplete because the options are missing; here are the options:

Which of the following is LESS dense than water?

The spoon

The glass

The tablets

The bubbles

The correct answer to this question is The bubbles

Explanation:

In general, the density of materials and substances affects their buoyancy. This implies in water less dense materials will float and those with higher density will sink. In the situation presented, the only element that is less dense than water are bubbles; this is shown by the movement of the bubbles as these originate in the bottom of the glass of water but they rise to the surface, which shows they are less dense than water.

6 0
3 years ago
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What volume of lead (of density 11.3 g/cm3 ) has the same mass as 395 cm3 of a piece of redwood (of density 0.38 g/cm3 )? Answer
Zepler [3.9K]
d_{1}=\frac{m}{V_{1}}\\\\
V_{1}=395cm^{3}\\
d_{1}=0,38\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ m=395cm^{3}*0,38\frac{g}{cm^{3}}=150,1g\\\\\\
d_{2}=\frac{m}{V_{2}}\\\\
m=150,1g\\
d_{2}=11,3\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ V_{2}=\frac{150,1g}{11,3\frac{g}{cm^{3}}}\approx13,28cm^{3}
8 0
3 years ago
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