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3 years ago

a nuclide of 64/29 cu absorbs a positron. witch is the resulting atom? (A) 65/29Cu (B) 63/29 CU (C) 64/28Ni (D) 64/30 Zn

Chemistry

1 answer:

Nesterboy [21]3 years ago

4 0

Answer : Option D) $Zn^{64}_{30}$

Explanation : When a positron is getting absorbed it means it will be $e^{+1}_{0}$ so, the $Cu^{64}_{29}$ will get converted;

So, the whole reaction will be;

$Cu^{64}_{29}$ + $e^{+1}_{0}$ ----> $Zn^{64}_{30}$ .

This will convert the whole element of Cu will get changed into Zn. As, it absorbs by the positron, the atomic number gets increased from 29 to 30.

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2 years ago

If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution

Helen [10]

This is a straightforward dilution calculation that can be done using the equation

$M_1V_1=M_2V_2$

where M₁ and M₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and V₁ and V₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (M₁) and the initial (V₁) and final (V₂) volumes, and we want to find the final concentration (M₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for M₂:

$M_2=\frac{M_1V_1}{V_2}.$

Substituting in our values, we get

$\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].$

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

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2 years ago

Which atoms in the amino acids become the h2o molecule produced by their action in model?

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1 year ago

1.00g of a metallic element reacts completely with 300cm3 of oxygen at 298K and 1 atm pressure to form an oxide which contains O

elena55 [62]

Answer:

The metal is most likely calcium which has a molar mass of 40.078 g

Note: The question is incomplete. The complete question is given below:

1.00g of a metallic element reacts completely with 300cm3 of oxygen at 298K and 1 atm pressure to form an oxide which contains O2– ions. The volume of one mole of gas at this temperature and pressure is 24.0dm3 .

What could be the identity of the metal?

A calcium B magnesium C potassium D sodium

Explanation:

Number of moles of the oxygen gas reacted is first determined:

Volume of oxygen gas = 300 cm³ = 0.30 dm³

1 mole of a gas occupies 24.0 dm³ at this temperature and pressure

Number of moles of oxygen gas that will occupy 0.30 dm³ = 0.3/24 = 0.125 moles

Let M be represent the metallic element. The equation of the reaction will be:

2M + O₂ ----> 2MO

From the equation of reaction;

2 moles of the metal reacts with 1 mole of oxygen gas

2 * 0.125 moles of the metal will react with 300 cm³ of oxygen = 0.025 moles

Since the mass of the metal that reacted with oxygen is 1.0 g, therefore, 1.0 g of the metal = 0.025 moles

mass of 1 mole of the metal = 1 /0.025 = 40.0 g

Therefore, the metal is most likely calcium which has a molar mass of 40.078 g

5 0

2 years ago