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MrRa [10]
3 years ago
9

A clear, pure liquid sample is brought into the lab and exposed to an electrical current. Different gases are

Chemistry
2 answers:
mel-nik [20]3 years ago
4 0
I think it’s a d but I’m not sure
Tresset [83]3 years ago
4 0
I think the answer is an element
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Does anyone know how to find mass of NaN3?
Alexus [3.1K]
The molar mass (atomic weight ) of sodium is 23.0 grams/mole and the molar mass of sodium azide, NaN3 , is the mass of sodium, 23.0 gram/mole added to the molar mass of three atoms of nitrogen (14.0 x 3 = 42 gram/mole) which equals 65.0 grams/mole. The percentage of sodium is 23.0 /65.0 x 100 % = 35 %
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3 years ago
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What is bonding all about in chemistry?
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A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds or through the sharing of electrons as in covalent bonds.
4 0
3 years ago
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7
Alenkasestr [34]

Answer:

B

Explanation:

Because the iron is wrong because of the contemplation of bisectors interfere with the question making it B.

3 0
3 years ago
Write an equation that represents the action in water of rubidium hydroxide as an Arrhenius base.
Anika [276]

Answer:

RbOH  → Rb⁺ +  OH⁻

As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

Explanation:

Arrhenius theory states that a compound is considered a base, if the compound can generate OH⁻ ions in aqueous solution.

Our compound is the RbOH.

When it is put in water, i can dissociate like this:

RbOH  → Rb⁺ +  OH⁻

As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

3 0
3 years ago
At high temperatures, carbon reacts with O2 to produce CO as follows: C(s) O2(g) 2CO(g). When 0.350 mol of O2 and excess carbon
777dan777 [17]

Answer:

Value of K_{c} is 0.090.

Explanation:

Initial molarity of O_{2} = \frac{0.350}{5.00}M = 0.0700 M

Construct an ICE table corresponding to the combustion reaction of carbon to determine K_{c}

                       C(s)+O_{2}(g)\rightarrow 2CO(g)

              I (M):    -      0.0700        0

             C (M):  -          -x             +2x

             E (M):   -    0.0700-x       2x

So, K_{c}=\frac{[CO]^{2}}{[O_{2}]}  , where [CO] and [O_{2}] represents equilibrium concentration of CO and O_{2} respectively.

Here, [CO]=2x=0.060

       ⇒x = 0.030

So, [O_{2}] = 0.0700-x = (0.0700-0.030) = 0.040

Hence,  K_{c}=\frac{(0.060)^{2}}{0.040}=0.090

4 0
4 years ago
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