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3 years ago

What is the mass of 3.0 x 10^23 atoms of neon

1 answer:

8 0

3.0e23 atoms Ne

"E" means 10^

Then we multiply it by a mole of Ne. By the definetion of a mole, it is always 6.022e23 atoms of an element.

So now, we do this:

3.0e23 atoms Ne x (1 mol Ne / 6.022e23 atoms Ne)

After that, we use molar mass. A mole of Neon is equal, in terms of grams, to its avg. atomic mass. This goes true for any element.

It ends up like this:

3.0e23 atoms Ne x (1 mol Ne / 6.022e23 atoms Ne) x (20.1797 g Ne / 1 mol Ne)

Now cancel out the "atoms Ne" and "1 mol Ne"

You end up with a grand total of...
*plugs everything into a calculator*

10.05298... g Ne.

We need to round to 2 sig. figs. (3.0) so now it's....

10 g Ne.

Note that this method can only be used for converting atoms of an element to mass in grams.
Source(s):
A periodic table for the atomic mass of neon.
A chemistry textboook
A chemistry class.

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Which property of matter is conserved in chemical reactions and shown by balanced equations? mass volume density shape

den301095 [7]

Answer:mass

Explanation:

I took the quiz

3 0

3 years ago

Read 2 more answers

One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) techniq

frozen [14]

Answer:

a. i. 8.447 × 10⁻³ T ii. 27.14 cm

b. i. 2.14 cm ii. It is easily detectable.

Explanation:

i. What strength of magnetic field is required?

Since the magnetic force F = Bqv equals the centripetal force F' = mv²/r on the C12 charge, we have

F = F'

Bqv = mv²/r

B = mv/re where B = strength of magnetic field, m = mass of C12 isotope = 1.99 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q = charge on C 12 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron)and r = radius of semicircle = 25.0 cm/2 = 12.5 cm = 12.5 × 10⁻² m

So,

B = mv/rq

B = 1.99 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (12.5 × 10⁻² m × 1.602 × 10⁻¹⁹ C)

B = 16.915 × 10⁻²³ kgm/s ÷ (20.025 × 10⁻²¹ mC)

B = 0.8447 × 10⁻² kg/sC)

B = 8.447 × 10⁻³ T

(ii) What is the diameter of the 13C semicircle?

Since the magnetic force F = Bq'v equals the centripetal force F' = mv²/r' on the C13 charge, we have

F = F'

Bq'v = mv²/r'

r' = mv/Be where r = radius of semicircle, B = strength of magnetic field = 8.447 × 10⁻³ T, m = mass of C12 isotope = 2.16 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q' = charge on C 13 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron) and = d/2 = 12.5 cm = 12.5 × 10⁻² m

So, r' = mv/Be

r' = 2.16 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (8.447 × 10⁻³ T × 1.602 × 10⁻¹⁹ C)

r' = 18.36 × 10⁻²³ kgm/s ÷ 13.5321 × 10⁻²² TC)

r' = 1.357 × 10⁻¹ kgm/TC)

r' = 0.1357 m

r' = 13.57 cm

Since diameter d' = 2r', d' = 2(13.57 cm) = 27.14 cm

i. What is the separation of the C12 and C13 ions at the detector at the end of the semicircle?

Since the diameter of the C12 isotope is 25.0 cm and that of the C 13 isotope is 27.14 cm, their separation at the end of the semicircle is 27.14 cm - 25.0 cm = 2.14 cm

ii. Is this distance large enough to be easily observed?

This distance of 2.14 cm easily detectable since it is in the centimeter range.

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2 years ago

Ca(OH)2 + H3PO4 = Ca3(PO4)2 + H2O

larisa [96]

Make a quick chart with each element represented, and count them up. HINT - leave the polyatomic anions together - in this case, PO4

Left Right
1 Ca 3
2 O 1
5 H 2
1 PO4 2

Begin by balancing like finding common denominators of fractions - apply to both sides:
I started by adding a 2 in front of H3PO4 on the left, them 6 in front of H2O on the right. Last, a 3 in front of Ca (OH)2. Then, re-count using the chart format to make sure you're right.

3Ca(OH)2 + 2H3PO4 = Ca3(PO4)2 + 6H2O

4 0

2 years ago

Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

$XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$ $\Delta H_1=-241.8kJ$

(2) $X(s)+2Cl_2(g)\rightarrow XCl_4(s)$ $\Delta H_2=+461.9kJ$

(3) $\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)$ $\Delta H_3=-92.3kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $H_2O(g)\rightarrow H_2O(l)$ $\Delta H_5=-44.0kJ$

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_1=2\times 241.8kJ=483.6kJ$