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yawa3891 [41]
3 years ago
15

What kind of intermolecular forces act between a hydrogen chloride molecule and a hydrogen iodide molecule?

Chemistry
1 answer:
wlad13 [49]3 years ago
5 0

Answer:

Since ΔEN > 0, the bond is covalent polar and the molecule is polar (dipole). Since ΔEN > 0, the bond is covalent polar and the molecule is polar (dipole). HI and ClF interact through a dipole-dipole force

Explanation:

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In the chemical equation Zn + 2HCl ZnCl2 + H2, the reactants are
Rudik [331]

Answer : The correct option is A.

Explanation :

The given chemical reaction is,

Zn+2HCl\rightarrow ZnCl_2+H_2

In the reaction, Reactants are present on the left side of the arrow and products are present on the right side of the arrow.

Reactants are those which takes part in and undergoes change during the reaction.

Products are those which are formed during the reaction.

In the given reaction, zinc react with hydrochloric acid to form zinc chloride and hydrogen gas.

Therefore, the reactants are zinc and hydrochloric acid.


4 0
3 years ago
Read 2 more answers
Convert the following Grams: 0.200 moles of H2S
ANEK [815]

Answer:

6.82 g H₂S

General Formulas and Concepts:

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

0.200 mol H₂S

<u>Step 2: Identify Conversions</u>

Molar Mass of H - 1.01 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

<u>Step 3: Convert</u>

  1. Set up:                    \displaystyle 0.200 \ mol \ H_2S(\frac{34.09 \ g \ H_2S}{1 \ mol \ H_2S})
  2. Multiply:                  \displaystyle 6.818 \ g \ H_2S

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

6.818 g H₂S ≈ 6.82 g H₂S

7 0
3 years ago
In order to produce an electric current in a coil of wire , a magnet field through the coil must be
IRINA_888 [86]

Answer:

magnet field must also be around the same coil to produce current

6 0
3 years ago
Read 2 more answers
A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential
nika2105 [10]

Answer:

3.50*10^-11 mol3 dm-9

Explanation:

A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential difference between the rod and the SHE is 0.5812 V, the rod being positive. Calculate the solubility product constant for silver oxalate.

Ag2C2O4 -->  2Ag+  +  C2O4 2-

So Ksp = [Ag+]^2 * [C2O42-]

In 1 L, 2.06*10^-4 mol of silver oxalate dissolve, giving, the same number of mol of oxalate ions, and twice the number of mol (4.12*10^-4) of silver ions.

So Ksp = (4.12*10^-4)^2 * (2.06*10^-4)

= 3.50*10^-11 mol3 dm-9

7 0
3 years ago
A. Gallium is produced by the electrolysis of a solution obtained by dissolving gallium oxide in concentrated NaOH(aq). Calculat
liraira [26]

Answer:

A)Mass of  gallium plated out is 0.3440 grams

B) For 0.67 hours current of 5.79 A must to be applied to plate out 8.70 g of tin.

Explanation:

To calculate the total charge, we use the equation:

C=I\times t

where,

C = Charge

I = Current in time t (seconds)

To calculate the moles of electrons, we use the equation:

\text{Moles of electrons}=\frac{C}{F}

where,

F = Faraday's constant = 96500

A) The equation for the deposition of Ga(s) from Ga(III) solution follows:

Ga^{3+}(aq.)+3e^-\rightarrow Ga(s)

I = 0.790 A, t = 30.0 min = 1800 seconds

C=I\times t

C=0.790 A\times 1800 s=1422 C

Moles of electron transferred:

=\frac{1422 C}{96500 F}=0.01474 mol

Now, to calculate the moles of gallium, we use the equation:

\text{Moles of Gallium}=\frac{\text{Moles of electrons}}{n}

n = number of electrons transferred = 3

\text{Moles of Gallium}=\frac{0.01474 mol}{3}=0.004913 mol

Mass of 0.004913 moles of gallium = 0.004913 mol × 70 g/mol=0.3440 g

B) The equation for the deposition of Sn(s) from Sn(II) solution follows:

Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)

Moles of tin = \frac{8.70 g}{119 g/mol}=0.07311 mol

n = number of electrons transferred = 2

\text{Moles of tin}=\frac{\text{Moles of electrons}}{n}

Moles of electron =  n\times \text{Moles of tin}

=2\times 0.07311 mol=0.14622 mol

Charge transferred during time t :

\text{Moles of electrons}=\frac{C}{F}

C=96500 F\times 0.14622 mol=14,110.23 C

Current applied for t time = I = 5.79 A

t=\frac{C}{I}=\frac{14,110.23 C}{5.79 A}=2,437 s=0.67 hrs

For 0.67 hours current of 5.79 A must to be applied to plate out 8.70 g of tin.

5 0
4 years ago
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