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2 years ago

Besides water, what is the product of a Neutalization Reaction between HBr and CsOH?

Chemistry

2 answers:

Luba_88 [7]2 years ago

5 0

I don’t know dawg but I’m just answering this so I can complete the steps good luck tho

BlackZzzverrR [31]2 years ago

3 0

Answer:

The answer to your question is the letter d. CsBr

Explanation:

Data

HBr reacts with CsOH

A neutralization reaction occurs between an acid and a base and the products will always be water and a binary salt.

In this reaction, the acid is HBr and the base is CsOH, then the products will be water and Cesium bromide.

Balanced chemical reaction

HBr + CsOH ⇒ CsBr + H₂O

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2 years ago

When a piece of metal is irradiated with UV radiation (λ = 162 nm), electrons are ejected with a kinetic energy of 3.54×10-19 J.

dsp73

We have that the work function of the metal

$\phi=1.227*10^{-18}J$

From the Question we are told that

UV radiation (λ = 162 nm)

Kinetic energy $K.E =3.54*10-19 J.$

Generally the equation for Kinetic energy is mathematically given as

$KE =\frac{hc}{\pi-\phi} \\\\\phi =\frac{ 6.626*10^{-34} * 3*10^8}{162*10^{-9} -3.54*10^{-19}}$

$\phi=1.227*10^{-18}J$

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2 years ago

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen

olga55 [171]

Answer: Rate law= $k[A]^1[B]^2$ , order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is $3L^2mol^{-2}s^{-1}$

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$Rate=k[A]^x[B]^y$

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: $1.2\times 10^{-2}=k[0.10]^x[0.20]^y$ (1)

From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (2)

Dividing 2 by 1 : $\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}$

$4=2^y,2^2=2^y$ therefore y=2.

b) From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (3)

From trial 3: $9.6\times 10^{-2}=k[0.20]^x[0.40]^y$ (4)

Dividing 4 by 3: $\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}$

$2=2^x,2=2^1$ , x=1

Thus rate law is $Rate=k[A]^1[B]^2$

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1: $1.2\times 10^{-2}=k[0.10]^1[0.20]^2$

$k=3 L^2mol^{-2}s^{-1}$ .

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3 years ago

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sweet [91]

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2 years ago

The density of silver is 10.49 g/cm3. if a sample of pure silver has a volume of 12.993 cm3, what would the mass?

weqwewe [10]

Multiply 10.49 by 12.993. that should be it. 130 grams ish?

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3 years ago