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jek_recluse [69]
3 years ago
14

A certain element forms an ion with 36 electrons and a charge of +2. identify the element. express your answer as a chemical sym

bol.
Chemistry
1 answer:
blsea [12.9K]3 years ago
6 0

The element is Sr (strontium)

strontium is in atomic number 38 in the periodic table. Strontium has 2.8.8.8.8.2

it loses two electrons to become  stable hence it has a charge of   2+. when strontium loses two electron it form ion with 36 electrons

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Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m
Nataly_w [17]
The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

<span>Trial      [A]      [B]         Rate </span><span>
<span>            (M)     (M)          (M/s) </span>
<span>1         0.50    0.010      3.0×10−3 </span>
<span>2         0.50    0.020       6.0×10−3 </span>
<span>3         1.00 0  .010       1.2×10−2</span></span>


Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0  =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

 
Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0   = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s
8 0
3 years ago
Read 2 more answers
you are given 500 mL of a 5M stock solution of ammonium chloride but for your experiment, you only need 100 mL of a 0.65M soluti
mafiozo [28]

Answer:

13ml

Explanation:

  1. to prepare dis solution first you need a volumetric flask of 100ml calibrated .
  2. using dilution formulae u will need 13ml from the stock
  3. measure 13ml of the 5M stock of the 500ml
  4. then drop it into the 100ml calibrated flask
  5. then add water till it reach d mark and you shake thoroughly
6 0
3 years ago
Solve for y in the following problem: 5.3 x 10- (y)(2y)
weqwewe [10]

Answer:

The value of y = 5.1478

Explanation:

The linear equation is an equation obtained when a linear polynomial is equated to zero. When the solution obtained on solving the equation is substituted in the equation in place of the unknown, the equation gets satisfied.

The given equation: 5.3 x 10- (y)(2y) = 0

⇒ 53 - 2y² = 0

⇒ 2y² = 53

⇒ y² = 53 ÷ 2 = 26.5

⇒ y = √26.5 = 5.1478

8 0
3 years ago
Hi everyone can anyone help with this, the question and diagram is in the pic thx!
seraphim [82]

Answer:

QP

Explanation:

P has 9 electrons.

Electronic Configuration : 2, 7

Valence electrons : 7

P needs 1 electron to get stable electronic configuration.

Q has 3 electrons.

Electronic Configuration : 2, 1

Valence electrons : 1

P needs to loose 1 electron to get stable electronic configuration.

Q donates 1 electron,

Q -----> Q+ + 1 e-

P gains 1 electron,

P + 1 e- -----> P-

Q+ + P- -----> QP

This is an ionic compound.

8 0
3 years ago
"What quantitative measurement and qualitative observations can be made about the interactions of matter and energy?" can someon
kirza4 [7]

Quantitative measurements are numerical values, they involve amounts and units like measuring things. Qualitative observations appeal to the five senses, like what does the interaction look and sound like

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3 years ago
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