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Sholpan [36]
2 years ago
13

7. Write the equation for the positron emission of barium-127.

Chemistry
1 answer:
ziro4ka [17]2 years ago
8 0

The reaction is given by

\\ \rm\Rrightarrow {}^{127}_{56}Ba\longrightarrow {}^{0}_{+1}\beta+{}^{127}_{55}Cs

Barium goes underneath beta decay to form Ceaseum

  • Cs is very mellable element
  • It can melt on your hand
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This flow chart shows the amount of energy that is emitted by each type of light. ultraviolet > blue light > yellow light
harina [27]

Answer:

Explanation:

In photoelectric effect , radiation of some energy is made to fall on metal plate which results in the ejection of electrons by the metal plate . The kinetic energy of electrons comes from the energy of radiation falling on metal plate . Some of energy of radiation falling on metal plate is used in the process of bringing the  electron to  the surface and it is called threshold energy . The radiation must have at least this energy to see to it that electrons are ejected .  and the rest of the energy of radiation is used in imparting kinetic energy to the  electron .

The red light radiation has least energy so it is least likely to eject electrons from metal plate and produce photoelectric effect .

3 0
3 years ago
Read 2 more answers
(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction
ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

Best regards!

5 0
3 years ago
Can you help me please?
almond37 [142]

Explanation:

Physical changes only change the appearance of a substance, not its chemical composition.

Chemical changes cause a substance to change into an entirely substance with a new chemical formula.

Chemical changes are also known as chemical reactions. The “ingredients” of a reaction are called reactants, and the end results are called products

7 0
3 years ago
Problem Page A chemist measures the amount of bromine liquid produced during an experiment. He finds that of bromine liquid is p
masha68 [24]

The question is incomplete, here is the complete question:

A chemist measures the amount of bromine liquid produced during an experiment. She finds that 766.g of bromine liquid is produced. Calculate the number of moles of bromine liquid produced. Round your answer to 3 significant digits.

<u>Answer:</u> The amount of liquid bromine produced is 4.79 moles.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given:

Given mass of liquid bromine = 766. g

Molar mass of liquid bromine, (Br_2) = 159.8 g/mol

Putting values in above equation, we get:

\text{Moles of liquid bromine}=\frac{766.g}{159.8g/mol}=4.79mol

Hence, the amount of liquid bromine produced is 4.79 moles.

6 0
3 years ago
A sample of gas occupies 9.0 mL at a pressure of 500.0 mm Hg. A new volume of the same sample is at a pressure of 750.0 mm Hg.
Anuta_ua [19.1K]
<h3>Answer:</h3>

                The New pressure (750 mmHg) is greater than the original pressure (500 mmHg) hence, the new volume (6.0 mL) is smaller than the original volume (9.0 mL).

<h3>Solution:</h3>

              According to Boyle's Law, " <em>The Volume of a given mass of gas at constant temperature is inversely proportional to the applied Pressure</em>". Mathematically, the initial and final states of gas are given as,

                                     P₁ V₁  =  P₂ V₂    ----------- (1)

Data Given;

                  P₁  =  500 mmHg

                  V₁  =  9.0 mL

                  P₂  =  750 mmHg

                  V₂  =  ??

Solving equation 1 for V₂,

                   V₂  =  P₁ V₁ / P₂

Putting values,

                   V₂  =  (500 mmHg × 9.0 mL) ÷ 750 mmHg

                   V₂  =  6.0 mL

<h3>Result:</h3>

            The New pressure (750 mmHg) is greater than the original pressure (500 mmHg) hence, the new volume (6.0 mL) is smaller than the original volume (9.0 mL).

4 0
4 years ago
Read 2 more answers
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