The molar mass of cu is 63.55 g mol. the number of grams of cu produced in this reaction is

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6. Given the following ' equation: Cu 2 AgNOs Cu(NOs) 2 Ag How many moles of Cu are needed to react with 3.50 moles of AgNO;? If 89.5 grams of Ag were produced, how many grams of Cu reacted?

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I am Hello. We need to find out the number of moles of the carpet. As we know, that is a balanced reaction. There is a silver nitrate. There is a balanced reaction for the silver nitrate and the CPU. That is a mole of copper. It is reacting with two moles of reactants and a piano. It was three. That is a ratio between this and another. 3.5 moles off ADA No three of AGN. It will be one divided by two and 3.5 mol of copper. This was corrected over the weekend, that is more of a G N03. 3.5 mol of a G N 03 is what it reacts with. It will be divided by two and then 3.5. There are hands on solving. 1.75 moles is the value that we will get. In the second part it is found that 89.5 g of silver is produced. Let's first find out the number of moles of silver which is producer and we know that it is equal to mass divided by moller Maas. Here is the mass given. The mass of silver is over 100 g per mole. Graham to Graham got canceled out. That is divided by 10. We will have 0.8 three moles. It is right there. We can say that the one mole of corporate is reacting and that it produced one mole of 80. So zero, that's right. This will be equal to a mole of silver, a mole of copper, and a mole of a couple. The mars that is equal to the number of moles, is there. The molar mass of a couple 0.8 reason number of moles and corporate is 63.547. To mole of AG is what we will get when we add these two values. It was produced by one mole. The bowl of copper will be divided by two, one by two and the other by 0.83 The value is 26.31 ground. I hope the concept is clear to you. Thanks so much.

Classification and Properties of Matter

In chemistry and physics, matter is any substance that has mass and takes up space by having volume. All everyday objects that can be touched are ultimately composed of atoms, which are made up of interacting subatomic particles, and in everyday as well as scientific usage, "matter" generally includes atoms and anything made up of them, and any particles and objects that act as if they have both rest mass and volume. However it does not include massless particles such as photons, or other energy phenomena or waves such as light or sound. Matter exists in various states (known as phases) that are defined by various physical properties, such as state of matter, phase, shape, and density. The Standard Model of particle physics and the general theory of relativity describe fundamental particles and the fundamental forces acting between them that control the structure and dynamics of matter.

Empirical Formula of a Hydrate

a
mass of the crucible (g)

88.000 g (0.5 pt)

b
mass of the crucible and hydrate (g)
93.000 g (0.5 pt)
c

mass of the crucible and pure salt (g)

91.196 g (0.5 pt)

Make sure to include your work for the following calculations.

a mass of water in the hydrate sample (g)

93.000 g - 91.196 g (0.5 pt) = 1.804 g (0.5 pt)

b number of moles of water in the sample

1.804 g x 1 mol water (1 pt) = 0.1 mol H2O (1 pt) 18 g H2O

c mass of pure salt in the sample (g)

5.000 g - 1.804 g (0.5 pt) = 3.196 g CuSO4 (0.5 pt)

d number of moles of salt in the sample

3.196 g x 1 mol CuSO4 (1 pt) = 0.02 mol CuSO4(1 pt) 159.55 g CuSO4

e molar ratio of water to salt in the sample

0.1 mol H2O (1 pt) = 5 mol H2O (1 pt) 0.02 mol CuSO4 1 mol CuSO4
Thus, 5:1

3. From your calculations, what is the empirical formula of copper sulfate hydrate?
4. Write the equation for the decomposition of the hydrate, CuSO4*XH2O.
CuSO4*5H2O (1.5 pts)
CuSO4*5H2OCuSO4 + 5H2O (1.5 pts)

Experiment 2
1. Record the following masses:
2. Calculate the following:
Make sure to include your work for the following calculations.

a
mass of the crucible (g)

88.000 g (0.5 pt)

b
mass of the crucible and hydrate (g)
93.000 g (0.5 pt)

c

mass of the crucible and pure salt (g)

90.342 g (0.5 pt)

a mass of water in the hydrate sample (g)

93.000 g - 90.342 g (0.5 pt) = 2.658 g (0.5pt)

b number of moles of water in the sample

2.658 g x 1 mol water (1 pt) = 0.148 mol H2O (1 pt) 18 g H2O

c mass of pure salt in the sample (g)

5.000 g - 2.658 g (0.5 pt) = 2.342 g MgCl2 (0.5 pt)

d number of moles of salt in the sample

2.342 g x 1 mol MgCl2 (1 pt) = 0.0246 mol MgCl2 (1 pt) 95.21 g MgCl2

e molar ratio of water to salt in the sample

0.148 mol H2O (1 pt) = 6 mol H2O (1 pt) 0.0246 mol MgCl2 1 mol MgCl2
Thus, 6:1

3. From your calculations, what is the empirical formula of magnesium chloride hydrate?
4. Write the equation for the decomposition of the hydrate, MgCl2*XH2O.
MgCl2*6H2O (1.5 pt)
MgCl2*6H2O -> MgCl2 + 6H2O (1.5 pt)

How do you convert g mol to grams?

How many grams are there in 3.2 moles of CU?

What is the molar mass of copper g mol?