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nataly862011 [7]
3 years ago
15

A 50.0 ml beaker contains 680. g mercury what is the density of mercury

Chemistry
1 answer:
slavikrds [6]3 years ago
5 0

Answer:

5.43 grams

Explanation:

because, Its one of the smallest planets, and is also very close to the Sun.

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The following is a sample question for brainliest. Brainliest will be given to the person who can answer the following question
Amanda [17]

Answer:

Chemistry is the branch of science that deals with the properties, composition, and structure of elements and compounds, how they can change, and the energy that is released or absorbed when they change. Traditionally, chemistry has been broken into five main subdisciplines: Organic, Analytical, Physical, Inorganic and Biochemistry. Chemistry is involved in everything we do. The reason why chemistry touches everything we do is because almost everything in existence can be broken down into chemical building blocks.

Explanation:

Hope this helps

8 0
3 years ago
What type of bond is sand
deff fn [24]

Answer:

Hello there! How are you doing today :] Sand would be a covalent bond.

Explanation:

Silica is one of the main compounds found in sand. It also contains many silicon and oxygen atoms. which; are joined together by covalent bonds in a regular arrangement, forming a giant covalent structure.

7 0
3 years ago
Which metal in Period 5 is very reactive and has two valence electrons in each atom?
Romashka-Z-Leto [24]
The answer is Strontium(Sr). The reactive increase from right to left. And this element has two valence electrons. So Rb is not correct. Then the very reactive metal is Sr.
8 0
3 years ago
Read 2 more answers
The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl
nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1  x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

3 0
3 years ago
You could add solid KCl to the solution to precipitate out AgCl(s). What mass of KCl is needed to precipitate the silver ions fr
padilas [110]

Answer:

0.143 g of KCl.

Explanation:

Equation of the reaction:

AgNO3(aq) + KCl(aq) --> AgCl(s) + KNO3(aq)

Molar concentration = mass/volume

= 0.16 * 0.012

= 0.00192 mol AgNO3.

By stoichiometry, 1 mole of AgNO3 reacts with 1 mole of KCl to form a precipitate.

Number of moles of KCl = 0.00192 mol.

Molar mass of KCl = 39 + 35.5

= 74.5 g/mol

Mass = molar mass * number of moles

= 74.5 * 0.00192

= 0.143 g of KCl.

4 0
3 years ago
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