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3 years ago

Which element has the electron configuration [Rn] 7s2 5f14 6d4 ?

Chemistry

2 answers:

nexus9112 [7]3 years ago

3 0

Answer:

Rn] 5f14 6d4 7s2 mp: none d: none. Seaborgium ..... Element. Density. Boiling or Melting Point in °C. Electron Configuration. Symbol.

Explanation:

solmaris [256]3 years ago

3 0

Answer: The element having given configuration is Seaborgium.

Explanation:

Electronic configuration tells us about the number of electrons that are present in an atom. It also determines the atomic number of an element.

We are given:

Electronic configuration = $[Rn]7s^25f^{14}6d^4$

Radon is the noble gas having atomic number 86. It has 86 number of electrons.

Total number of electrons in the given element = (86 + 2 + 14 +4) = 106

Element having atomic number '106' is Seaborgium

Hence, the element having given configuration is Seaborgium.

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A series of six solutions was prepared at the given concentrations. Sample A B C D E F Concentration (mM) 50 100 150 200 300 400

ioda

Answer:

300 mM

Explanation:

In order to solve this problem we need to calculate the line of best fit for those experimental values. The absorbance values go in the Y-axis while the concentration goes in the X-axis. We can calculate the linear fit using Microsoft Excel using the LINEST function (alternatively you can write the Y data in one column and X data in another one, then use that data to create a dispersion graph and finally add the line of best fit and its formula).

The formula for the line of best fit for this set of data is:

y = 0.005 * x

So now we calculate the value of x when y is 1.50:

1.50 = 0.005 * x

x = 300 mM

7 0

3 years ago

What is the pH of a solution with a POH of 8.35?

Paha777 [63]

Answer: The pH of the solution is 5.65

Explanation:

The relationship between the pH and the pOH is that $pH+pOH=14$ .

Given this, we can plug in the pOH and subtract that from 14.

$14-pOH=pH\\14-8.35=5.65$

I hope this helped! Pls give brainliest!! :)

3 0

2 years ago

For a reaction A + B → products, the following data were collected. Experiment Number Initial Concentration of A (M) Initial Con

Afina-wow [57]

Answer:

Rate constant k = 1.57*10⁻⁵ s⁻¹

Explanation:

Given reaction:

$A\rightarrow B$

Expt [A] M [B] M Rate [M/s]

1 3.40 4.16 1.82*10^-4

2 4.59 4.16 3.32*10^-4

3. 3.40 5.46 1.82*10^-4

$Rate = k[A]^{x}[B]^{y}$

where k = rate constant

x and y are the orders wrt to A and B

To find x:

Divide rate of expt 2 by expt 1

$\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2$

To find y:

Divide rate of expt 3 by expt 1

$\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0$

Therefore: x = 2, y = 0

$Rate = k[A]^{2}[B]^{0}$

To find k

Use rate for expt 1:

$k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1$

6 0

2 years ago

Elements and compounds are two types of what

fredd [130]

Elements and compounds are two types of substances ( I think). An element is any material made up of one type of atom- there could be multiples of this atom though for example oxygen which has a formula of 02. A compound is a substance made up of more than one type of atom for example H2O Hope this helped! :)

7 0

3 years ago

Carbonic anhydrase of erythrocytes (Mr 30,000) has one of the highest turnover numbers known. It catalyzes the reversible hydrat

Afina-wow [57]

Explanation:

According to the given data, the turnover number can be calculated as follows.

Turnover number = $K_{cat} = \frac{V_{max}}{\text{Concentration of enzyme}}$

$V_{max} = \frac{\text{Moles of CO_{2} hydrolyzed}{second}$

Therefore, moles of $CO_{2}$ hydrolyzed is as follows.

Moles of $CO_{2}$ hydrolyzed = $\frac{Mass of CO_{2}}{Molar mass of CO_{2}}$

= $\frac{0.30}{44}$

= 0.00682 moles

Now, moles of $CO_{2}$ hydolyzed per second is calculated as follows.

Moles of $CO_{2}$ hydolyzed per second = $\frac{0.00682}{60}$

= $1.137 \times 10^{-4} moles/second = V_{max}$

And,

Moles of enzyme = $\frac{Mass}{\text{Molar mass}}$

= $\frac{10.0 \mu g}{30000}$

= $3.33 \times 10^{-10}$ moles

Therefore, the value of $K_{cat}$ is as follows.

$K_{cat} = \frac{1.137 \times 10^{-4} moles}{3.333 \times 10^{-10} moles}$

= $0.3411 \times 10^{6}$ per second

= $0.3411 \times 60 \times 10^{6}$ per minute

= $20.466 \times 10^{6}$ per minute

Thus, we can conclude that the turnover number ( $K_{cat}$ ) of carbonic anhydrase (in units of $min^{-1}$ ) is $20.466 \times 10^{6}$ per minute.

6 0

3 years ago