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2 years ago

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Enter the expression 2gΔym−−−−√, where Δ is the uppercase Greek letter Delta. Note: the term Δy represents a single variable, no

t two separate variables multiplied together.

1 answer:

Hatshy [7]2 years ago

5 0

Answer:

Square root { 2g * Delta-y * /m}

Explanation:

it is as shown in the attached file.

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Q 11.20: What is the product of the reaction between t-BuCl and MeOH? A : t-BuOH B : MeOCl C : t-BuOMe D : (CH3)2CCH2

Sholpan [36]

Answer:

C : t-BuOMe

Explanation:

The tert -butanol is a tertiary alcohol and when chloride ion attacks the carbocation, it forms t-BuCl.

The reaction of tert-butyl chloride or t-BuCl ((CH3)3C−Cl) with methanol and MeOH (CH3−OH) gives the product tert-Butyl methyl ether or t-BuOMe (CH3)3C−OCH3:

(CH3)3C−Cl + CH3−OH => (CH3)3C−OCH3 + HCl

Hence, the correct asnwer is C : t-BuOMe

5 0

2 years ago

A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t

TEA [102]

Explanation:

The reaction equation will be as follows.

$Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)$

So, according to this equation, 1 mole $Ca(OH)_{2}$ = 2 mol HBr = 1 mol $CaBr_{2}$

Therefore, calculate the number of moles of calcium hydroxide as follows.

No. of moles of $Ca(OH)_{2}$ = $V \times Molarity$

= $50 \times 0.6$

= 30 mmol

Similarly, calculate the number of moles of HBr as follows.

No. of moles of HBr = $M \times V$

= $50 \times 0.6$

= 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of $CaBr_{2}$ = $30 \times \frac{1}{2}$

= 15 mmol

Hence, calculate the amount of heat released as follows.

Heat released in the reaction(q) = $m \times s \times \Delta T$

as, m = mass of solution

and, Density = $\frac{mass}{volume}$

or, mass = Density × Volume

= 1.08 g/ml \times (50 + 50) ml

= 108 g

where, s = specific heat of solution = 4.18 j/g.k

and, change in temperature $\Delta T$ = $(26 - 23)^{o}C$

= $3
^{o}C$

Hence, the heat released will be as follows.

q = $m \times s \times \Delta T$

q = $108 \times 4.18 \times 3^{o}C$

= 1354.32 joule

or, = 1.354 kJ (as 1 kJ = 1000 J)

Also, $\Delta H_{rxn}$ = $\frac{-q}{n}$

= $\frac{-1.354}{15 \times 10^{-3}}$

= -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

6 0

2 years ago

The enthalpy of formation of water is –285.8 kJ/mol. What can be inferred from this statement?

LiRa [457]

A negative formation enthalpy means that the reaction is exothermic, or that heat is released during the process.
(C,)

4 0

2 years ago

Read 2 more answers

The volume of a gas is 7.15 l measured at 1.00 atm. what is the pressure of the gas in mmhg if the volume is changed to 9.25 l?

IRINA_888 [86]

1 atm=7.15/9.25
Volume increase comes from reduced pressure

3 0

2 years ago

Which of the following statements concerning hydrocarbons is/are correct?

Alona [7]

Answer:

1. Saturated hydrocarbons may be cyclic or acyclic molecules.

2. An unsaturated hydrocarbon molecule contains at least one double bond.

Explanation:

Hello,

In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).

Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true

Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.

Therefore, first and second statements are correct.

Best regards.

6 0

2 years ago