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N76 [4]
3 years ago
9

Iodine would have chemical properties most like -

Chemistry
1 answer:
MaRussiya [10]3 years ago
3 0
Iodine would have chemical properties most like H. chlorine.
They would have similar chemical properties because they belong to the same group in the periodic system of elements, unlike manganese, tellurium, and xenon. To see which group these belong to, just look at the periodic system vertically. 

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The original concentration in a sample is 5.1 x 10^9 cfu/ml. which sample volume should yield a countable plate
poizon [28]

The original concentration is 5.1 × 10⁹ CFU / ml

In order to attain a countable plate, the number of CFU must be present in between 10 and 200 per ml.

Let us take 0.1 ml and dilute it to 1 ml.

This minimizes the concentration to 5.1 × 10⁹ × 10⁻¹ = 5.1 × 10⁸ CFU/ml

In order to minimize the concentration in between 10 and 200, it can be reduced to 5.1 × 10¹

The final concentration = 5.1 × 10¹ CFU/ml

Initial concentration = 5.1 × 10⁹ CFU/ml

Volume of sample with 5.1 × 10¹ CFU = 5.1 × 10¹ CFU × (1 ml / 5.1 × 10⁹ CFU)

= 1.0 × 10⁻⁸ ml

This is the volume to be taken to obtain countable value, 51 CFU.

6 0
3 years ago
HELP plz<br> Are beaches made from erosion or deposition?
soldier1979 [14.2K]

Answer:

Desposition

Explanation:

7 0
3 years ago
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Atoms of which of the following elements are largest?<br> A:<br> Na<br> OOOO
elena-14-01-66 [18.8K]

Answer:

yes answer os Na because it's electronic configuration is 1s^2,2s^2,2p^6,3s^1

5 0
3 years ago
You are asked to make 500 mL of a 0. 250 M sodium chloride (NaCl) solution.
nignag [31]

Answer:

a. .125 mol b. 7.25g

Explanation:

a. To get the moles multiply the volume by the molarity.

.5L*.25M=.125mol NaCl

b. The molar mass is provided so multiply the moles by the molar mass to get the amount of grams of nacl.

.125mol*58molar mass=7.25g nacl to make the solution

3 0
2 years ago
Given the following balanced equation at 120°C: A(g) + B(g) ⇋ 2 C(g) + D(s)(a) At equilibrium a 4.0 liter container was found to
BlackZzzverrR [31]

Answer:

a) kc = 0,25

b) [A] = 0,41 M

c) [A] = <em>0,8 M</em>

[B] =<em>0,2 M</em>

[C] = <em>0,2M</em>

Explanation:

The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.

Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium.

If global reaction is:

A(g) + B(g) ⇋ 2 C(g) + D(s)

The kc = \frac{[C]^2}{[A][B]}

a) The concentrations of each compound are:

[A] = \frac{1,60 moles}{4,0 L} = <em>0,4 M</em>

[B] = \frac{0,40 moles}{4,0 L} = <em>0,1 M</em>

[C] = \frac{0,40 moles}{4,0 L} = <em>0,1 M</em>

<em>kc = </em>\frac{[0,1]^2}{[0,4][0,1]} = 0,25

b) The addition of B and D in the same amount will, in equilibrium, produce these changes:

[A] = \frac{1,60-x moles}{4,0 L}

[B] = \frac{0,60-x moles}{4,0 L}

[C] = \frac{0,60+2x moles}{4,0 L}

0,25 = \frac{[0,60+2x]^2}{[1,60-x][0,60-x]}

You will obtain

3,75x² +2,95x +0,12 = 0

Solving

x =-0,74363479081119   → No physical sense

x =-0,043031875855476

Thus, concentration of A is:

\frac{1,60-(-0,04 moles)}{4,0 L} = <em>0,41 M</em>

c) When volume is suddenly halved concentrations will be the concentrations in equilibrium over 2L:

[A] = \frac{1,60 moles}{2,0 L} = <em>0,8 M</em>

[B] = \frac{0,40 moles}{2,0 L} = <em>0,2 M</em>

[C] = \frac{0,40 moles}{2,0 L} = <em>0,2M</em>

I hope it helps!

8 0
4 years ago
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