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3 years ago

¿soy guapo? si dices que si te doy una galletita

Chemistry

2 answers:

tankabanditka [31]3 years ago

8 0

Answer:

Hindi ko po ma gets sinasabi nyu pwedeng pakiayus

bezimeni [28]3 years ago

3 0

Chrehtegv ittgger ok so eva

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Calculate the energy required to heat 322.0g of ethanol from −2.2°C to 19.6°C . Assume the specific heat capacity of ethanol und

just olya [345]

Answer:

There is 17.1 kJ energy required

Explanation:

Step 1: Data given

Mass of ethanol = 322.0 grams

Initial temperature = -2.2 °C = 273.15 -2.2 = 270.95K

Final temperature = 19.6 °C = 273.15 + 19.6 = 292.75 K

Specific heat capacity = 2.44 J/g*K

Step 2: Calculate energy

Q = m*c*ΔT

⇒ m = the mass of ethanol= 322 grams

⇒ c = the specific heat capacity of ethanol = 2.44 J/g*K

⇒ ΔT = T2 - T1 = 292.75 - 270.95 = 21.8 K

Q = 322 * 2.44 * 21.8 = 17127.8 J = 17.1 kJ

There is 17.1 kJ energy required

3 0

3 years ago

Which conclusion is supported by the information in the graph?

Molodets [167]

Answer:

A. The cost of producing a kilowatt of power with fuel cell will be less than $30 in 2015

4 0

2 years ago

The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10

Sati [7]

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0 125 kPa

At t = teq 125 - 2x 4x x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $2.8\times 10^{-3}$ min⁻¹

Initial concentration $[A_0]$ = 125 kPa

Final concentration $[A_t]$ = 91 kPa

Time = ?

Applying in the above equation, we get that:-

$91=125e^{-2.8\times 10^{-3}\times t}$

$125e^{-2.8\times \:10^{-3}t}=91$

$-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)$

$t=113\ min$

3 0

3 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

3 years ago

Which of the following is the correct chemical formula for cs and br? csbr cs2br csbr2

nalin [4]

The correct chemical formulae is CsBr

4 0

3 years ago

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