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2 years ago

An incorrect way of handling disposable gloves includes

Engineering

1 answer:

Alexxandr [17]2 years ago

4 0

The incorrect way of handling disposable gloves are as follows;

Blow into them
Hold them by the fingers
Roll them onto the hands

<h3>What are disposable gloves?</h3>

Disposable gloves are gloves that you use one time and then throw away.

They are usually transparent and made of nitrile, latex, or vinyl.

They are usually tough to withstand liquids and chemicals.

Therefore, the incorrect way of handling them includes the following;

Blow into them
Hold them by the fingers
Roll them onto the hands

learn more on disposable gloves here: brainly.com/question/24012826

#SPJ11

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The reverse water-gas shift (RWGS) reaction is an equimolar reaction between CO2 and H2 to form CO and H2O. Assume CO2 associati

klasskru [66]

Answer:

a) Check explanation for this

b)Rate law is $Rate = \frac{k_{1}k_{4} }{k_{3}+ 2k_{4} } [H_{2} ]$

c) The rate does not depend on the concentration of CO₂

Explanation:

a) Elementary steps for the RWGS reaction:

Dissociative adsorption of the H₂ Molecule

$H_{2} $\xrightarrow{\text{k1}}$H + H$ (Fast process)

Reversible Reaction between CO₂ and H

$CO_{2} + H\mathrel{\mathop{\rightleftarrows}^{\mathrm{k2}}_{\mathrm{k3}}}COOH$ (Fast Process)

Slow dissociation of COOH into gaseous CO and absorbed OH

$COOH $\xrightarrow{\text{k1}}$ CO + OH$ (Slow process)

Fast hydrogenation of the OH to form H₂O

$OH + H $\xrightarrow{\text{k5}}$H_{2} O$ (Fast process)

b) Derivation of the rate law

We need to determine the rate law for H, OH and COOH because these are the intermediates for this reaction.

The steady state approximation is applied to a consecutive reaction with a slow first step and a fast second step (k1≪k2). If the first step is very slow in comparison to the second step, there is no accumulation of intermediate product.

Rate of consumption = Rate of production

For COOH:

Using steady state approximation

$\frac{d[COOH]}{dt} = 0$

$k_{2} [CO_{2} ][H] = k_{3} [COOH] k_{4} [COOH]\\$

$[COOH] = \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4} } \\$

For H:

$\frac{d[H]}{dt} = 0$

$k_{1}[H_{2}] = k_{2}[CO_{2} [H]+k_{5} [ OH][H]$

$[H]= \frac{k_{1}[H_{2}] }{k_{5}[OH] +k_{2}[CO_{2}]}\\$

For OH:

$\frac{d[OH]}{dt} = 0$

$k_{4} [COOH] = k_{5} [OH][H]\\\k[OH] = \frac{k_{4} [COOH]}{k_{5} H}\\$

The rate of the overall reaction is determined by the slowest step of the reaction. The slowest process is the dissociation of COOH

Therefore the overall rate of reaction is:

$Rate = k_{4} [COOH]\\$

$Rate = k_{4} \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4} }\\Rate = k_{4} \frac{k_{2}[CO_{2}]\frac{k_{1}[H_{2}] }{k_{5}[OH] +k_{2}[CO_{2}]} }{k_{3}k_{4}}\\Rate = k_{4} \frac{k_{2}[CO_{2}]\frac{k_{1}[H_{2}] }{k_{5}\frac{k_{4}COOH }{k_{5}H } +k_{2}[CO_{2}]} }{k_{3}k_{4}}$

Simplifying the equation above, the rate law becomes

$Rate = \frac{k_{1}k_{4} }{k_{3}+ 2k_{4} } [H_{2} ]$

c) It is obvious from the rate law written above that the rate of the RWBG reaction does not depend on the concentration of CO₂

7 0

3 years ago

Can you carry 1 m3 of liquid water? Why or why not? (provide the weight to support your answer)

djverab [1.8K]

Answer:

No we cannot carry 1 cubic meter of liquid water.

Explanation:

As we know that density of water is 1000 kilograms per cubic meter of water hence we infer that 1 cubic meter of water will have a weight of 1000 kilograms of 1 metric tonnes which is beyond the lifting capability of strongest man on earth let alone a normal human being who can just lift a weight of 100 kilograms thus we conclude that we cannot lift 1 cubic meter of liquid water.

5 0

3 years ago

In 2009 an explosive eruption covered the island of Hunga Ha'apai in black volcanic ash. What type of succession is this?

dangina [55]

Answer:

The type of succession is:

Primary succession

Explanation:

This is a type of succession that occurs after a volcanic eruption or earthquake; it involves the breakdown of rocks by lichens to create new, nutrient rich soils.

Primary succession is one of the two types of succession we have. It begins on rock formations, such as volcanoes or mountains, or in a place with no organisms or soil.

8 0

3 years ago

Electricians will sometimes call ______ "disconnects" or a "disconnecting means."

Natali5045456 [20]

I want to say D but i’m not 100% sure

5 0

3 years ago

g A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner

galben [10]

This question is not complete, the complete question is;

A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner surface and oriented normal to the hoop stress direction. Repeated pressure cycling enabled the crack to grow larger. If the fracture toughness of the material is $88 Mpam^\frac{1}{2}$ , the yield strength equal to 1250 MPa, and the hoop stress equal to 300 MPa, would the vessel leak before it ruptured

Answer:

length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

Explanation:

Given the data in the question;

vessel thickness = 6 cm

fracture toughness k = $88 Mpam^\frac{1}{2}$

yield strength = 1250 MPa

hoop stress equal = 300 MPa

we know that, the relation between fracture toughness and crack length is expressed as;

k = (1.1)(2/π)(r√(πa))

where k is the fracture toughness, r is hoop stress and a is length of crack

so we rearrange to find length of crack

a = 1/π[( k / 1.1(r)(2/π)]²

a = 1/π[( kπ / 1.1(r)(2)]²

so we substitute

a = 1/π [( 88π / 1.1(300)(2/π)]²

a = 1/π[ 0.1754596 ]

a = 0.05585 m

a = 0.05585 × 100 cm

a = 5.585 cm

so, length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

8 0

3 years ago