All categories

3 years ago

Which of the following statements best describes the properties of aluminum-28?

Chemistry

1 answer:

Contact [7]3 years ago

5 0

d. When aluminum-28 undergoes beta decay, silicon-28 is produced.

Explanation:

When the nuclei of aluminium-28 decays, it produces silicon- 28:

Aluminium ²⁸₁₃Al

Silicon 28 ²⁸₁₄Si

beta particle ⁰₋₁ $\beta$

²⁸₁₃Al → ²⁸₁₄Si + ⁰₋₁ $\beta$

This way, the mass and atomic number are conserved.

Conservation of mass number:

28 = 28 + 0, 28 = 28

13 = 14 -1 , 13 = 13

Learn more:

Balancing nuclear equations brainly.com/question/10094982

#learnwithBrainly

You might be interested in

Which body system is this? Your brain and nerves receive and send

KiRa [710]

Answer: its the nervous system

Explanation:

8 0

3 years ago

A gas has a pressure of 0.470 atm at 60.0 C. What is the pressure at standard temperature?

OLEGan [10]

Answer:

P2=0.385atm

Explanation:

step one:

Given that the temperature T1= 60 Celcius

we can convert this to kelvin by adding 273k to 60 Celcius

we have T1= 333k

pressure P1= 0.470 atm

step two:

we know that the standard temperature is T2= 273K

Applying the temperature and pressure relationship we have

P1/T1=P2/T2

substituting our given data we have

0.47/333=P2/273

cross multiply we have

P2= (0.47*273)/333

P2= 128.31/333

P2=0.385 atm

6 0

3 years ago

What is one bad thing that can happen when humans build more and more streets and buildings?

Irina18 [472]

Answer:

there will be less and less space

Explanation:

4 0

2 years ago

For the reaction CO+2H2=CH3OH at 700 K, equilibrium concentrations are [H2]=0.072 M, [CO]= 0.020M, and [CH3OH]= 0.030 M. Calcula

bekas [8.4K]

The balanced equation for the reaction is

CO(g) + 2H₂(g) ⇄ CH₃OH(g)

The given concentrations are at equilibrium state. Hence we can use them directly in calculation with the expression for the equilibrium constant, k. expression for k can be written as

k = [CH₃OH(g)] / [CO(g)] [H₂(g) ]²

[H₂]=0.072 M
[CO]= 0.020M
[CH₃OH]= 0.030 M

From substitution,

k = 0.030 M / 0.020 M x (0.072 M)²

k = 289.35 M⁻²

Hence, equilibrium constant for the given reaction at 700 K is 289.35 M⁻².

3 0

3 years ago

ou will prepare 250-mL of this solution using a 30% (m/v) NaOH stock solution. How many mL of the NaOH stock solution will you n

ArbitrLikvidat [17]

Answer:

$\boxed{\text{3.3 mL}}$

Explanation:

You must convert 30 % (m/v) to a molar concentration.

Assume 1 L of solution.

1. Mass of NaOH

$\text{Mass of NaOH} = \text{1000 mL solution } \times \dfrac{\text{30 g NaOH}}{\text{100 mL solution}} = \text{300 g NaOH}$

2. Moles of NaOH

$\text{Moles of NaOH} = \text{300 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40.00 g NaOH}} = \text{7.50 mol NaOH}$

3. Molar concentration of NaOH

$c= \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{7.50 mol}}{\text{1 L}} = \text{7.50 mol/L}$

4. Volume of NaOH

Now that you know the concentration, you can use the dilution formula .

$c_{1}V_{1} = c_{2}V_{2}$

to calculate the volume of stock solution.

Data:

c₁ = 7.50 mol·L⁻¹; V₁ = ?

c₂ = 0.1 mol·L⁻¹; V₂ = 250 mL

Calculations:

(a) Convert millilitres to litres

$V = \text{250 mL} \times \dfrac{ \text{1 L}}{\text{1000 mL}} = \text{0.250 L}$

(b) Calculate the volume of dilute solution

$\begin{array}{rcl}7.50V_{1} & = & 0.1 \times 0.250\\7.50V_{1} &= & 0.0250\\V_{1} & = & \text{0.0033 L}\\& = & \textbf{3.3 mL}\\\end{array}$

$\text{You will need $\boxed{\textbf{3.3 mL}}$ of the stock solution.}$

4 0

2 years ago