All categories

3 years ago

Isotopes of the same element always have the same (3 points)

1 answer:

8 0

Isotopes of the same element always have the same atomic number because the atomic number is like the identity of each element and if it is changed the element is changed for example isotopes of Carbon are: ₆C¹² , ₆C¹³ and ₆C¹⁴ and if we change the atomic number form 6 to 7 we will have Nitrogen element ₇N¹⁴
So the correct answer is Z- number (atomic number)

You might be interested in

When blood passes through the lungs, carbon dioxide and water are formed from carbonic acid dissolved in the blood. The carbon d

NeTakaya

Answer is: decomposition.

Balanced chemical reaction: H₂CO₃ → CO₂ + H₂O.
H₂CO₃ is carbonic acid.
CO₂ is carbon (IV) oxide or carbon dioxide.
Chemical decomposition is the separation of a single chemical compound (in this example carbonic acid) into its two or more simpler compounds (in this example water and carbon dioxide).

7 0

4 years ago

Read 2 more answers

What are easy ways to Study Chemistry ??

Elina [12.6K]

Answer:

seneca

past papers

take notes from videos

Explanation:

very good website, asks questions about the subject correct for your exam board and gives correct answers and explanations

exam papers always help

7 0

3 years ago

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500=193000C$ is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

$14\times 1000g=14000g$ of copper is deposited by = $\frac{193000}{63.5}\times 14000=42551181 C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

$40.0=\frac{42551181 C}{t}\\\\t=1063779sec$

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, $1063779s\times \frac{1hr}{3600s}=295hr$

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0

4 years ago

How many grams of aluminum oxide (Al2O3) will be formed from 10.0 grams of aluminum (Al)? 4 AL + 3 02 --> 2 AL203

d1i1m1o1n [39]

Answer:

Mass = 18.9 g

Explanation:

Given data:

Mass of Al₂O₃ formed = ?

Mass of Al = 10.0 g

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 10.0 g/ 27 g/mol

Number of moles = 0.37 mol

Now we will compare the moles of Al and Al₂O₃.

Al : Al₂O₃

4 : 2

0.37 : 2/4×0.37 = 0.185 mol

Mass of Al₂O₃:

Mass = number of moles × molar mass

Mass = 0.185 mol × 101.9 g/mol

Mass = 18.9 g

4 0

3 years ago

Ne, c4h10, is a component of natural gas that is used as fuel for cigarette lighters. the balanced equation of the complete comb

Simora [160]

3.68 liters First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements: Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286 Looking at the balanced equation for the reaction which is 2 C4H10(g)+13 O2(g)â†’8 CO2(g)+10 H2O(l) It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have: 0.037851286 mol * 4 = 0.151405143 mol The ideal gas law is PV = nRT where P = Pressure V = Volume n = number of moles R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) ) T = absolute temperature (23C + 273.15K = 296.15K) So let's solve the formula for V and the calculate using known values: PV = nRT V = nRT/P V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm) V = (3.679338871 L*atm)/(1 atm) V = 3.679338871 L So the volume of CO2 produced will occupy 3.68 liters.

4 0

3 years ago