Từ một dung dịch có pH=7 muốn tạo thành dung dịch có pH nhỏ hơn 7 thì cho vào dung dịch

What happens to electrons in the photoelectric effect?

mafiozo [28]

The correct answer is B. on Apex!

8 0

3 years ago

Read 2 more answers

Mr. Ragusa asks Hassan to make silver crystals from the following reaction.

SVEN [57.7K]

Answer:

Percentage yield = 61.7%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated:

2AgNO₃ + Cu —> Cu(NO₃)₂ + 2Ag

Next, we shall determine the mass of AgNO₃ that reacted and the mass of Ag produced from the balanced equation. This is illustrated below:

Molar mass of AgNO₃ = 108 + 14 + (16×3)

= 108 + 14 + 48

= 170 g/mol

Mass of AgNO₃ from the balanced equation = 2 × 170 = 340 g

Molar mass of Ag = 108 g/mol

Mass of Ag from the balanced equation = 2 × 108 = 216 g

SUMMARY:

From the balanced equation above,

340 g of AgNO₃ reacted to produce 216 g of Ag.

Next, we shall determine the theoretical yield of Ag. This can be obtained as follow:

From the balanced equation above,

340 g of AgNO₃ reacted to produce 216 g of Ag.

Therefore, 51 g of AgNO₃ will react to produce = (51 × 216)/340 = 32.4 g of Ag.

Thus, the theoretical yield of Ag is 32.4 g.

Finally, we shall determine the percentage yield of Ag. This can be obtained as follow:

Actual yield = 20 g

Theoretical yield = 32.4 g

Percentage yield =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 20 / 32.4 × 100

Percentage yield = 61.7%

6 0

3 years ago

How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?

xenn [34]

Answer:

$5.225x10^{22}atoms\ I$

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

$n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2$

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

$atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I$

Best regards!

4 0

3 years ago

What is the percent composition by mass of oxygen in Ca(NO3)2 (gram-formula= 164 g/mol)?

Neko [114]

To find this, we will use this formula:

Molar mass of element
------------------------------------ x 100
Molar mass of compound

So, first lets calculate the mass of the compound as a whole. We use the atomic masses on the periodic table to determine this.

Ca: 40.078 g/mol
N2 (there is two nitrogens): 28.014 g/mol
O6 (there are six nitrogens: 3 times 2): 95.994 g/mol

When we add all of those numbers up together, we get 164.086. That is the molar mass for the whole compound. However, we are trying to figure out what percent of the compound oxygen makes up. From the molar mass, we know that 95.994 of the 164.086 is oxygen. Lets plug those numbers into our equation!

95.994
-----------
164.086

When we divide those two numbers, we get .585. When we multiply that by 100, we get 58.5.

So, the percent compostition of oxygen in Ca(NO3)2, or, calcium nitrate, is 58.5%.

5 0

3 years ago

: Citric acid, H3C6H5O7, is a triprotic acid. Consider a buffer system comprising H2C6H5O7 - and HC6H5O7 2- ions. What is the ne

olchik [2.2K]

Answer:

The Net reaction is