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Radda [10]
3 years ago
13

Describing physical and chemical changes:

Chemistry
1 answer:
svetoff [14.1K]3 years ago
8 0
Physical: The chemist could try to bend it to find out how malleable it is. He could also try to pull it into wires to find out how ductile it is.

Chemical: The chemist could put the metal into contact with other substances to get an idea of how reactive it is, and he could try to burn it and find out how flammable it is.
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The gas usually filled in the electric bulb is
Mariulka [41]

Answer:

1st answer:  A. nitrogen

2nd answer:  A. Sodium carbonate

Explanation:

8 0
3 years ago
Read 2 more answers
Write the names and symbols for four elements in each of the following categories: (a) nonmetal, (b) metal, (c)metalloid.
Over [174]

Explanation:

Non-metals are the species that are electron deficient and they are able to accept one or more electrons from a donor atom in order to complete their octet.

For  example, carbon (C), nitrogen (N), chlorine, (Cl), phosphorus (P) etc are all non-metals.

Metals are the species that contain more number of electrons in their valence shell and in order to attain stability they easily lose an electron.

For example, sodium (Na), lithium (Li), Beryllium (Be), Magnesium (Mg) etc are all metals.

Metalloids are the species that show properties of both metals and non-metals.

For example, Boron (B), Antimony (Sb), Silicon (Si) and Germanium (Ge) etc are metalloids.

5 0
3 years ago
I will mark brainliest
Veronika [31]

I believe the answer is B??????????? Hope this helps

~Queensupreme

5 0
3 years ago
Read 2 more answers
Science question down below
aliya0001 [1]
Option 1/A (It is the first one)
6 0
3 years ago
Yet a third pair of compounds of manganese and oxygen is 50.48% and 36.81% oxygen respectively. In what small whole number ratio
Mariulka [41]

Answer:

The number ratio is 4:7

Explanation:

Step 1: Data given

Compound 1 has 50.48 % oxygen

Compound 2 has 36.81 % oxygen

Molar mass oxygen = 16 g/mol

Molar mass manganese = 54.94 g/mol

Step 2: Calculate % manganes

Compound 1: 100 - 50.48 = 49.52 %

Compound 2: 100 - 36.81 = 63.19 %

Step 3: Calculate mass

Suppose mass of compounds = 100 grams

Compound 1:

 50.48 % O = 50.48 grams

 49.52 % Mn = 49.52 grams

Compound 2:

36.81 % O = 36.81 grams

63.19 % Mn = 63.19 grams

Step 4: Calculate moles

Compound 1

Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles

Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles

Compound 2

Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles

Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles

Step 5: calculate mol ratio

We will divide by the smallest amount of moles

Compound 1

O: 3.155/0.9013 = 3.5

Mn: 0.9013 / 0.9013 = 1

Mn2O7

Compound 2

O: 2.301 / 1.150 = 2

Mn: 1.150 / 1.150 = 1

MnO2

The number ratio is 2:3.5 or 4:7

7 0
3 years ago
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