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Dmitrij [34]
3 years ago
14

How many atoms of phosphorus are in 3.80 mol of copper(II) phosphate?

Chemistry
1 answer:
user100 [1]3 years ago
3 0
The amount of atoms of Phosphorus in 3.80 mol of copper (II) phosphate would be equivalent with :
3.80 mol x (6.022 x 10^23 molecules)/ (1 mol)

Hope this helps
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Which of the following is NOT true about a C-14 atom?
romanna [79]

Answer:

option C. it contain 6 neutrons.

Explanation:

7 0
3 years ago
In everyday speech the words precision and accuracy are often used interchangeably. When these terms are used in science are the
padilas [110]

Answer:

            No, in science their meanings are not the same as their everyday meanings.

Explanation:

                   In Science, Precision and Accuracy are defined as,

Accuracy:

               Accuracy is the value which is closest to the known or standard value.

Precision:

                While, Precision is the value of closeness of two measured values to each other.

Example:

             Let suppose in Chemistry Lab you weight an object as 50 g. While the actual weight of that object is 30 g. It means your reading is not accurate.

             On second measurement you find that the object weight is 31 g. This time your reading is not precise.

3 0
4 years ago
Is anybody good in Chemistry that can help me with this?​
sesenic [268]

Answer:

just replace the 9 mole with 3.68 g of Al .

I think it will help you.

5 0
3 years ago
Select the correct answer
Assoli18 [71]

Answer:

B. an element

Explanation:

An atom is smallest indivisible particle that takes part in a chemical reaction. Different atoms due to the number of their protons called atomic number gives an element. Every element is a singular atom on it's own. Combination of atoms leads to the formation of molecules and compounds.

When compounds mix together without an actual chemical change, a mixture forms.

Elements are distinct substances that cannot be split up into simpler substances. They are usually made up of only one kind of atom.

4 0
3 years ago
n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains
Nitella [24]

Answer : The metal used was iron (the specific heat capacity is 0.449J/g^oC).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of unknown metal = ?

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of unknown metal = 150 g

m_2 = mass of water = 200 g

T_f = final temperature of water = 34.3^oC

T_1 = initial temperature of unknown metal = 150.0^oC

T_2 = initial temperature of water = 25.0^oC

Now put all the given values in the above formula, we get

150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC

c_1=0.449J/g^oC

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is 0.449J/g^oC).

6 0
4 years ago
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