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3 years ago

What is the atomic size going across a period (left or right)? what causes this trend?

1 answer:

5 0

The atomic size decreases; this is because as you go across the period, more electrons are found in the outer shell. As more electrons are added, the more attraction there is between the protons, in the nucleus, and the electrons, making the atom smaller.
Hope this helps!

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a student is holding a model of the human brain whht are two benefits of this model? side note: HELP IT'S A TEST AND I ONLY HAVE

mojhsa [17]

Answer:

One, It can help children to easily learn or recognize the structure of the brain.

Second, It can help children to learn about the ridges and groves, blood vessels in the brain which will grow curiosity in them to do more research about the stucture and function of the brain.

Explanation:

I hope this helps :)

6 0

3 years ago

g The decomposition reaction of A to B is a first-order reaction with a half-life of 2.42×103 seconds: A → 2B If the initial con

Tanzania [10]

Answer:

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Explanation:

The equation used to calculate the constant for first order kinetics:

$t_{1/2}=\frac{0.693}{k}}$ .....(1)

Rate law expression for first order kinetics is given by the equation:

$t=\frac{2.303}{k}\log\frac{[A_o]}{[A]}$ ......(2)

where,

k = rate constant

$t_{1/2}$ =Half life of the reaction = $2.42\times 10^3 s$

t = time taken for decay process = ?

$[A_o]$ = initial amount of the reactant = 0.163 M

[A] = amount left after time t = 66.8% of $[A_o]$

[A]= $\frac{66.8}{100}\times 0.163 M=0.108884 M$

$k=\frac{0.693}{2.42\times 10^3 s}$

$t=\frac{2.303}{\frac{0.693}{2.42\times 10^3 s}}\log\frac{0.163 M}{0.108884 M}$

t = 1,409.19 s

1 minute = 60 sec

$t=\frac{1,409.19 }{60} min=23.49 min$

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

6 0

3 years ago

What acid is car batteries?

Romashka-Z-Leto [24]

Answer:

sulfuric acid

Explanation:

When a battery is being discharged, the sulfuric acid in the electrolyte is being depleted so that the electrolyte more closely resembles water.

5 0

4 years ago

Read 2 more answers

Excess oxygen gas (O2) reacts with 244g of Iron (Fe) to produce 332 g of Fe2O3. What is the percent yield?

Andru [333]

Answer:

95.15%

Explanation:

To calculate the percent yield, we need the following formula:

$\% yield=\dfrac{actual yield}{theoreticalyield}\times100\%$

Solving for the theoretical yied, you need to predict how much of the product will be produced if we USE up the given.

Our given is 224g of Fe and we to get the theoretical yield, we need to figure out how much product will Fe produce supposing that we use up all the reactant.

First thing we do is get the balance equation of this chemical reaction:

4Fe + 3O₂ → 2Fe₂O₃

We get the ratio between Fe and the the product, Fe₂O₃

$\dfrac{4moles of Fe}{2molesofFe_{2}O_{3}}=\dfrac{2moles of Fe}{1moleofFe_{2}O_{3}}$

This basically means that we need 2 moles of Fe to produce 1 mole of Fe₂O₃. We'll use this later.

Now we let's use our given:

We need to first convert our given to moles. To do this, we need to determine how many grams there are of the reactant for every mole. We need to first get the atomic mass of the elements involved in the substance:

Iron(1)

Fe = 55.845(1) = 55.845g/mole(1)

Then we use this to convert grams to moles

$244g\times\dfrac{1mole}{55.845g}=4.369moles$

This means that there are 4.396moles of Fe in 244g of Fe.

This we will use to see how many moles of product we can produce given the moles of reactant by using the reactant:rproduct ratio.

$4.369moles of Fe\times\dfrac{1moleofFe_{2}O_{3}}{2molesofFe}=2.185 moles of Fe_{2}O_{3}$

So given 4.693 moles of Fe we can produce 2.185 moles of Fe₂O₃

The next step is to get how many grams of product there are given our calculation. We do this again by getting how many grams of Fe₂O₃ there are in 1 mole.

Fe(2) O(3)

Fe₂O₃=55.845(2) + 15.999(3)

= 111.69 + 47.997 =159.687g/mol

We then use this to solve for how many grams of product there are in 2.185 moles.

$2.185moles\times\dfrac{159.687g}{1mole}=348.92g$

This is our theoretical yield 348.92g of Fe₂O₃.

We can finally use our percent yield equation. Our actual yield is given by the probelm, 332g of Fe₂O₃ and we solved for our theoretical yield which is 348.92g of Fe₂O₃. We plug this in our formula and solve.

$\%yield=\dfrac{actual yield}{theoreticalyield}\times100\%$