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nordsb [41]
2 years ago
13

3. Mercury is a liquid metal that has a density of 13.58 g/mL. Calculate the volume of mercury

Chemistry
1 answer:
Alona [7]2 years ago
5 0

Answer:

0.03682 mL of mercury

Explanation:

We know the density of the mercury which is 13.58 g/mL

density = mass / volume

volume = mass / density

Now we can calculate the volume of 0.5 g of mercury:

volume = 0.5 / 13.58 = 0.03682 mL of mercury

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Imagine that you are given the mass spectra of these two compounds, but the spectra are missing the compound names.
12345 [234]

The structures of the isomers and the m/z values of their peaks are not given in the question. The complete question is provided in the attachment

Answer:

Compound 2 (2,5-dimethylhexane) will not have the peaks at 29 and 85 m/z

Explanation:

The fragmentation of molecules by electron ionization of mass spectrometer occurs according to Stevenson's Rule, which states that "The most probable fragmentation is the one that leaves the positive charge on the fragment with the lowest ionization energy". This is much like the Markovnikov's Rule in organic chemistry which has predicted the formation of most stable carbocation and the addition of hydrogen halide to it.

The mass spectra of compound 1 (2,4-dimethylhexane) will contain all the m/z values mentioned in the question. Each peak indicate towards homologous series of fragmentation product of the compound 1. The first peak can be attributed to ethyl carbocation (m/z = 29), with the increase of 14 units the next peak indicates towards propyl carbocation (m/z = 43) and onwards until molecular ion peak of 114 m/z.

Compound 2 (2,5-dimethylhexane) structure shows that the cleavage  of C-C bond will not yield a stable ethyl and hexyl carbocation. Hence, no peaks will be observed at 29 and 85 m/z. The absence of these two peaks can be used to distinguish one isomer from the other.

5 0
2 years ago
Exactly 17.0 mL of a H2SO4 solution was required to neutralize 45.0 mL of 0.235 M NaOH. What was the concentration of the H2SO4
aleksandr82 [10.1K]

Answer:

Molarity for the sulfuric acid is 0.622 M

Explanation:

When we neutralize an acid with a base, molarity of both . both volume are the same. The formula is:

M acid . volume of acid = M base . volume of base

M acid = unknown

Volume of acid = 17 mL

Volume of base = 45 mL

M base = 0.235 M

Therefore, we replace:  M acid . 17 mL = 0.235 M . 45 mL

M acid = (0.235 M . 45 mL) / 17 mL

M acid = 0.622 M

6 0
2 years ago
A 126-gram sample of titanium metal is heated from 20.0°C to 45.4°C while absorbing 1.68 kJ of heat. What is the specific heat o
Radda [10]

Answer:

The specific heat for the titanium metal is 0.524 J/g°C.

Explanation:

Given,

Q = 1.68 kJ   = 1680 Joules

mass = 126 grams

T₁ = 20°C

T₂ = 45.4°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

Here, ΔT =  T₂ - T₁ = 45.4 - 20 = 25.4°C.

Substituting values,

1680 = (126)(25.4)(Cp)

By solving,

Cp = 0.524 J/g°C.

The specific heat for the titanium metal is 0.524 J/g°C.

3 0
2 years ago
Which statement best describes the relationship between igneous and metamorphic rocks?
solong [7]

Answer:

Igneous rocks must go through the sedimentary process to change into metamorphic rocks. Igneous rocks are chemically changed into metamorphic rocks because of high temperature and pressure. Metamorphic rocks are formed from melting igneous rocks. Metamorphic rocks and igneous rocks do not follow a rock cycle.

Explanation:

hope this helps!

6 0
2 years ago
Read 2 more answers
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (
djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

Kb = \frac{[ HSO3-][0H-]}{SO3-2}

Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation Kb=\frac{[HSO3-][OH-]}{[SO3-2]}

We noe have:

1.485*10^-^7=\frac{x*x}{(0.056-x)}

x = [OH-] = 9.11*10^-^5

pOH = -log(OH) = -log(9.11*10^-^5)

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0
2 years ago
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