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2 years ago

How do chemists use systems and order in their work?

Chemistry

1 answer:

san4es73 [151]2 years ago

3 0

Answer:

5 point) Chemists used systems and order because they study the basic unit of all forms of matter, which is the atom . However, there are several different kinds of atoms, called elements . Chemists organize these elements using the periodic table .

Explanation:

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Please help Which of the following is the SI unit used to measure length? a.kilogram b.liter c.meter d.Kelvin

Oksanka [162]

<h3>Answer :</h3>

A. kilogram is the SI unit used to measure mass of body.

1 kilogram = 100 gram

B. litre is the SI unit used to measure volume of substance.

1 litre = 10ˉ³ m³

C. meter is the unit used to measure length of body.

1 meter = 100 centimeters

D. Kelvin is the unit used to measure temperature of body.

1 K = -273.15 °C

Hence, (B) is the correct answer!

Cheers!

6 0

3 years ago

Mixture formation is a ________change.

xxMikexx [17]

Chemical should be your answer.

6 0

3 years ago

What mass of iron (III) nitrate will be in 129.8ml of a 0.3556 molar aq iron (III) nitrate solution?

Anvisha [2.4K]

Answer: The mass of iron (III) nitrate is 11.16 g/mol

Explanation:

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.3556 M

Molar mass of Iron (III) nitrate = 241.86 g/mol

Volume of solution = 129.8 mL

Putting values in above equation, we get:

$0.3556M=\frac{\text{Mass of iron (III) nitrate}\times 1000}{241.86 g/mol\times 129.8}\\\\\text{Mass of iron (III) nitrate}=11.16g$

Hence, the mass of iron (III) nitrate is 11.16 g/mol

6 0

3 years ago

A binary compound of boron and hydrogen has the following percentage composition: 78.14% boron, 21.86% hydrogen. If the molar ma

algol [13]

Answer:

Empirical formula: BH3

Molecular Formula: B2H6

Explanation:

To solve the exercise, we need to know how many boron atoms and how many hydrogen atoms the compound has. We know that of the total weight of the compound, 78.14% correspond to boron and 21.86% to hydrogen. As the weight of the compound is between 27 g and 28 g, using the above percentages we can solve that the compound has between 21.1 g and 21.8 g of boron, and between 5.9 g and 6.1 g of hydrogen:

100% _____ 27 g

78.14% _____ x = 78.14% * 27g / 100% = 21.1 g boron

100% ______27 g

21.86% ______ x = 21.86% * 27g / 100% = 5.9 g hydrogen

100% _____ 28 g

78.14% _____ x = 78.14% * 28g / 100% = 21.8 g boron

100% _____ 28g

21.86% _____ x = 21.86% * 28g / 100% = 6.1 g hydrogen

So, if the atomic weight of boron is 10.8 g, there must be two boron atoms in the compound that sum 21.6 g. The weight of hydrogen is 1 g, so the compound must have six hydrogen atoms.

The molecular formula represents the real amount of atoms that form a compound. Therefore, the molecular formula of the compound is B2H6.

The empirical formula is the minimum expression that represents the proportion of atoms in a compound. For example, ethane has 2 carbon atoms and 6 hydrogen atoms, so its molecular formula is C2H6, however, its empirical formula is CH3. Therefore, the empirical formula of the boron compound is BH3.

8 0

3 years ago

Will hydrogen ever make a double bond?

Andrei [34K]

No, hydrogen can only hold one bond and that's it. It only needs to be paired with one bond.

7 0

3 years ago

Read 2 more answers