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SOVA2 [1]
2 years ago
8

What frequency does television channel 7 broadcast at?

Chemistry
2 answers:
lana [24]2 years ago
8 0

Answer:

Channel 7's frequency is 117.5 MHz

scZoUnD [109]2 years ago
6 0

Answer:

174 MHz - 180 MHz w/ center frequency of 177 MHz

Explanation:

216 - 174 = 42

42/7 = 6

6 + 174 = 180          -           7                   -               177 MHz

180 + 6 = 186          -           8                   -               183 MHz

186 + 6 = 192          -           9                   -               189 MHz

192 + 6 = 198          -          10                   -               195 MHz

198 + 6 = 204          -          11                   -               201 MHz

204 + 6 = 210          -          12                  -               207 MHz

210 + 6 = 216          -           13                  -               213 MHz

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Covalent bonding is important between atoms of biological molecules, but other factors are important for self-assembly. For exam
Gekata [30.6K]

Answer:

1. Polar

2 nonpolar

3. Nitrogen or oxygen

4.carbon

5.Repelled by and attracted to

5 0
3 years ago
Calculate the solubility of CaF2 in g/L (Ksp = 4.0 x 10-8). 2. What is the pH of a solution containing a hydrogen ion concentrat
Pepsi [2]

Answer:

\large \boxed{1. \text{ 0.17 g/L; 2. 3.52; 3. Cl; 4. (a) +3; (b) +4; (c) +6}}

Explanation:

1. Solubility of CaF_2

(a) Molar solubility

CaF₂ ⇌ Ca²⁺ + 2F⁻

K_{\text{sp }} = \text{[Ca$^{2+}$]}\text{[F$^{-}$]}^{2}= 4.0 \times 10^{-8}\\s(2s)^{2}=4.0 \times 10^{-8}\\4s^{3} = 4.0 \times 10^{-8}\\s^{3} = 1.0 \times 10^{-8}\\s =2.2 \times 10^{-3}\text{ mol/L}

(b) Mass solubility

\text{Solubility} = 2.2 \times 10^{-3} \text{ mol/L} \times \dfrac{\text{78.07 g}}{\text{1 L }} = \text{0.17 g/L}\\\\\text{The solubility of CaF$_{2}$ is $\large \boxed{\textbf{0.17 g/L}}$}

2. pH

pH = -log [H⁺] = -log(3.0 × 10⁻⁴) = 3.52

3. Oxidizing and reducing agents

Zn + Cl₂ ⟶ ZnCl₂

\rm \stackrel{\hbox{0}}{\hbox{Zn}} + \stackrel{\hbox{0}}{\hbox{ Cl}_{2} }\longrightarrow \stackrel{\hbox{+2}}{\hbox{Zn}}\stackrel{\hbox{-1}}{\hbox{Cl}_{2}}

The oxidation number of Cl has decreased from 0 to -1.

Cl has been reduced, so Cl is the oxidizing agent.

4. Oxidation numbers

(a) Al₂O₃

\stackrel{\hbox{$\mathbf{+3}$}}{\hbox{Al}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{3}}

1O = -2; 3O = -6; 2Al  = +6; 1Al = +3

(b) XeF₄

\stackrel{\hbox{$\mathbf{+4}$}}{\hbox{Xe}}\stackrel{\hbox{-1}}{\hbox{F}_{4}}

1F = -1; 4F = -4; 1 Xe = +4

(c) K₂Cr₂O₇

\stackrel{\hbox{${+1}$}}{\hbox{K}_{2}}\stackrel{\hbox{$\mathbf{+6}$}}{\hbox{Cr}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{7}}

1K = +1; 2K = +2; 1O = -2; 7O = -14

+2 - 14 = -12

2Cr = + 12; 1 Cr = +6

8 0
3 years ago
Copper(II) sulfate pentahydrate, CuSO4 ·5 H2O, (molar mass 250 g/mol) can be dehydrated by repeated heating in a crucible. Which
prohojiy [21]

Answer:

The water lost is 36% of the total mass of the hydrate

Explanation:

<u>Step 1:</u> Data given

Molar mass of CuSO4*5H2O = 250 g/mol

Molar mass of CuSO4 = 160 g/mol

<u>Step 2:</u> Calculate mass of water lost

Mass of water lost = 250 - 160 = 90 grams

<u>Step 3:</u> Calculate % water

% water = (mass water / total mass of hydrate)*100 %

% water = (90 grams / 250 grams )*100% = 36 %

We can control this by the following equation

The hydrate has 5 moles of H2O

5*18. = 90 grams

(90/250)*100% = 36%

(160/250)*100% = 64 %

The water lost is 36% of the total mass of the hydrate

8 0
2 years ago
A 0.300 m solution of hcl is prepared by adding some 1.50 m hcl to a 500 ml volumetric flask and diluting to the mark with deion
liberstina [14]
In dilution we add distilled water to decrease the concentration of required sample from high concentration to lower concentration
The law used for dilution:
M₁V₁]Before dilution = M₂V₂] After dilution
M₁ = 1.5 M
V₁ = ?
M₂ = 0.3 M
V₂ = 500 ml
1.5 * V₁ = 0.3 * 500 ml
so V₁ = 100 ml and it completed to 500 ml using 400 ml deionized water
8 0
3 years ago
Why saturated fatty acid can’t create kink ?​
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Saturated fatty acids exhibit a linear structure while unsaturated fatty acids bend, or kink, due to double bonds within the chemical foundation.
7 0
3 years ago
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