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IRINA_888 [86]
3 years ago
11

Suggest the conditions needed for hydrogen and iodine to react. Give reasons for your answer.

Chemistry
1 answer:
fenix001 [56]3 years ago
7 0
HI can be prepared by combing H2 and I2. This reaction is used to generate high-purity samples. This mixture of gases is being irradiated with the wavelength of light equal to the dissociation energy of I2, about 578 nm. This supports a mechanism whereby I2 first dissociates into 2 iodine atoms, which each attach themselves to a side of an H2 molecule and break the H−H bond.
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43 Which element reacts spontaneously with 1.0M HCl (aq) at room temperature?
KonstantinChe [14]
The correct answer is (4) zinc~

According to the reactivity series of metals, only zinc react with HCl (aq) at room temperature.

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4 0
3 years ago
A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the
Tresset [83]

Answer:

a. Molarity= M =2.1x10^{-1}M

b. Molality= m=2.0x10^{-1}m

Explanation:

Hello,

In this case, given the information about the aniline, whose molar mass is 93g/mol, one could assume the volume of the solution is just 200 mL (0.200 L) as no volume change is observed when mixing, therefore, the molarity results:

M=\frac{n_{solute}}{V_{solution}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L} =2.1x10^{-1}M

Moreover, the molality:

m=\frac{n_{solute}}{m_{solvent}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L*\frac{1.05kg}{1L} } =2.0x10^{-1}m

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7 0
3 years ago
Given the following sets of values, calculate the unknown
weqwewe [10]

Answer:

3.91 L

Explanation:

Using the ideal gas law equation as follows:

PV = nRT

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P = pressure (atm)

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n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

Based on the information given in this question,

P = 5.23 atm

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n = 0.831 mol

T = 27°C = 27 + 273 = 300K

Using PV = nRT

V = nRT/P

V = (0.831 × 0.0821 × 300) ÷ 5.23

V = 20.47 ÷ 5.23

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8 0
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Yuki888 [10]
The Answer To This Is B.
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rosijanka [135]

Answer:

my answer is b oti d I am not sure

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