How many electrons can the px atomic orbital subtype hold?

Help me asappp please

Basile [38]

Answer:

wait

Explanation:

wait

3 0

3 years ago

10 atoms of aluminum reacts with 6 molecules of oxygen gas to produce aluminum oxide. 4 Al + 3 O2 ––> 2 Al2O3 What is the exc

Mrrafil [7]

The excess reactant is Aluminum.

<u>Explanation:</u>

We have to write the balanced equation as,

4 Al+ 3 O₂ → 2 Al₂O₃

According to the molar ratio 4: 3, from the given balanced equation, we can say that 4 atoms of Al reacted with 3 molecules of oxygen.

Given that 10 atoms of aluminum reacts with 6 molecules of oxygen, as per the ratio only 8 atoms of Aluminum is required to react with 6 molecules of oxygen, so excess reactant is Aluminum.

8 0

3 years ago

What is the average yearly rate of change of carbon-14 during the first 5000 years?

erica [24]

Answer:

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year

Explanation:

Given that the mass of the carbon 14 at the start = 5 gram

At the end of 5,000 years we will have;

$A = A_0 \times e^{-\lambda \times t}$

Where

A = The amount of carbon 14 left

A₀ = The starting amount of carbon 14

e = Constant = 2.71828

$T_{1/2}$ = The half life

$\lambda = 0.693/T_{1/2}$

t = The time elapsed = 5000 years

λ = 0.693/ $T_{1/2}$ = 0.693/5730 = 0.0001209424

Therefore;

A = 5 × e^(-0.0001209424×5000) = 2.7312 grams

Therefore, the amount of carbon 14 decayed in the 5000 years is the difference in mass between the starting amount and the amount left

The amount of carbon 14 decayed = 5 - 2.7312 = 2.2688 grams

The average yearly rate of change of carbon-14 during the first 5000 years is therefore;

2.2688 grams/(5000 years) = 0.0004538 grams per year

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year.

5 0

3 years ago

How does the composition of water molecule affect its charge

kolezko [41]

It is make the molecule polar.

7 0

3 years ago

Read 2 more answers

A. Calculations for the Determination of Ammonium Chloride The data from the data entry portion of the report has been copied in

Vladimir79 [104]

Answer:

A

Explanation:

Considering question A

Mass of original sample is $m_o = 0.945 \ g$

Mass of NH4Cl is $m_n = 0.116 \ g$

Percent of NH4Cl is $k = 12.275 \%$

B

Mass of NaCl $m_k = 0.359 \ g$

C

Mass of SiO2 $m_e = 0.46$

D

Mass of original sample $m_o = 0.945 \ g$

Differences in these weights (g) (use the absolute value of the difference)

recovery of matter $G = 0.01 \ g$

The correct option is C

From the question we are told that

The mass of evaporating dish on #1 is $m_1 = 38.646 \ g$

The mass of evaporating dish and original sample $m_2 = 39 591 \ g$

The mass of evaporating dish after subliming $NH_4Cl$ is $m_3 = 39.4750 \ g$

Generally the mass of the original sample is mathematically represented as

$m_o = m_2 - m_1$

=> $m_o = 39 591 - 38.646$

=> $m_o = 0.945 \ g$

Generally the mass of $NH_4Cl$ is mathematically represented as

$m_n = m_2 - m_3$

=> $m_n = 39 591 - 39.4750$

=> $m_n = 0.116 \ g$

The Percent $NaH_4 Cl (g)$

$k = \frac{ m_n}{m_o} *100$

=> $k = \frac{0.116 }{0.945} *100$

=> $k = 12.275 \%$

Considering question B

The mass of evaporating dish #2 is $m_g = 38700\ g$

The mass of watch glass is $m_a = 28 299 \ g$

The mass of evaporating dish #2, watch glass and NaCl $m_b = 67,355 \ g$

Generally the mass of NaCl is

$m_k = m_b -[m_g + m_a]$

=> $m_k = 67,355 -[38700 + 28 299]$

=> $m_k = 0.359 \ g$

Considering question C

The mass of evaporating dish is $m_p= 38.645$

The mass of evaporating dish and SiO2 $m_s = 39.105 \ g$

Generally the mass of SiO2 is mathematically represented as

$m_e = 39.105 - 38.645$

=> $m_e = 0.46$

Considering D

The mass of the original sample is $m_o = 0.945 \ g$

Generally the experimental mass recovered (NH_4Cl,NaCl, SiO2 ) is mathematically evaluated as

$M =0.116 + 0.46 + 0.359$

$M = 0.935 \ g$

Generally the differences in these weights (g) of recovery of matter is mathematically represented as.

$G =0.945- 0.935$

=> $G = 0.01 \ g$

While drying the NaCl, the liquid boiled and some splattered out of the evaporating dish, causing the recovered mass to be lower.

6 0

3 years ago