Block A is made of copper, which has a specific heat value of 0.385 J/g*C.

BRAINLIEST ASAP!!! Help

Andru [333]

Answer:

N2C14

Explanation:

determined the bond type by looking if it is a metal or nometal

Ionic Bond:NM+M

Covalent Bond:NM+NM

7 0

3 years ago

Calculate the number of O atoms in 0.364 g of CaSO4 · 2H2O

Nikolay [14]

Answer:

= 7.66 x 10²¹ oxygen atoms in 0.364 grams of CaSO₄·2H₂O

Explanation:

For problems like this posting, one needs an understanding of the following topics:

The definition of the mole

1 mole of substance = mass in grams of substance containing 1 Avogadro's Number ( = 6.023 x 10²³ ) of particles of the specified substance. This is generally one formula weight of the substance of interest. From this, the following equivalent relationships should be memorized:

1 mole = 1 formula weight = 1 mole weight (g)= 6.023 x 10²³ particles

Converting grams to moles:

Given grams => moles = grams/gram formula wt

Converting moles to grams:

Given moles => grams = moles x gram formula wt

_________________________________________________________

Calculate the number of O atoms in 0.364 g of CaSO₄ · 2H₂O.

Given mass CaSO₄ · 2H₂O = 0.364 grams

Formula Wt CaSO₄ · 2H₂O = 172 g/mole

moles CaSO₄ · 2H₂O = mass CaSO4 · 2H2O / formula Wt. CaSO₄ · 2H₂O

= 0.364 g CaSO₄·2H₂O / 172 g CaSO4·2H2O

= (0.364/172) mole CaSO₄·2H₂O

= 2.12 x 10⁻³ mole CaSO₄·2H₂O

∴ number of Oxy (O) atoms in 0.364 grams CaSO₄·2H₂O

= (2.12 x 10⁻³ mole CaSO₄ · 2H₂O)(6.023 x 10²³ molecules CaSO₄· 2H₂O/ mole)

= 1.276876 x 10²¹molecules CaSO₄· 2H₂O CaSO₄2H₂O

= 1.276876 x 10²¹ molecules CaSO₄· 2H₂O x 6 oxygen atoms / molecule

= 7.661256 x 10²¹ oxygen atoms in 0.364 grams of CaSO₄·2H₂O

= 7.66 x 10²¹ oxygen atoms in 0.364 grams of CaSO₄·2H₂O

8 0

2 years ago

How many moles of H atoms are in 15.2 grams of pure ice?

nikitadnepr [17]

The number of moles of H atoms in 15.2 grams of pure ice is 0.850 moles

The number of moles of O atoms in 15.2 grams of pure ice is 13.50 moles

An atom is a smallest indivisible particle of a chemical element that is capable of independent existence.

Pure ice contains the H2O molecules and the molar mass of H2O is 18.02 g/mol

Using the relation:

number of moles = mass/molar mass

The number of moles of H2O = 15.2 grams/18.02 g/mol

number of moles of H2O = 0.8435 moles

If the atomic mass of H and O are as follow:

the atomic mass of H atom is = 1.00784 g/mol
the atomic mass of O atom is = 15.999 g/mol

Then:

the number of H atoms in 15.2 grams of pure ice = mass of H atom × molar mass of H2O.

the number of H atoms = 0.8435 mol × 1.00784 g/mol

the number of H atoms = 0.850 moles

the number of O atoms in 15.2 grams of pure ice = 0.8435 mol × 15.999 g/mol

the number of O atoms = 13.50 moles

Learn more about atoms here:

brainly.com/question/25832904

6 0

2 years ago

Which of these is not a precaution taken at nuclear power plants to ensure safety? Safety systems are built into them. All parts

irga5000 [103]

Answer: Enriched uranium is the fuel source.

The nuclear power plants utilizes the radiation energy from the fission of the heavy molecular weight atomic particles. These radiations can cause damage when exposed to the external air, they are capable of causing explosion, mutation in the living beings. Therefore, safety measures should be taken to prevent the damages.

Enriched uranium is the fuel source is not the precaution taken at the nuclear power plant to ensure safety because it is just explaining the fact that the uranium is the source of radiation. All other factors such as safety systems are built into them, all parts are kept in good repair and processes are carefully monitored are relevant precautions.

8 0

3 years ago

A hydraulic system is designed to lift cars for inspection in a service station. The narrow end of the system has a surface area

fiasKO [112]

Given:

Surface area at the narrow end, A1 = 5.00 cm2

Force applied at the narrow end, F1 = 81.0 N

Surface area at the wide end, A2 = 725 cm2

To determine:

Force F2 applied at the wide end

Explanation:

Use the relation

F1/A1 = F2/A2

F2 = F1*A2/A1 = 81.0 N * 725 cm2/5.00 cm2 = 11,745 N

Ans: (b)

The force applied at the wide end = 11,745 N

5 0

2 years ago

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