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3 years ago

During hydrolysis, ___________ must be added before the bonds can be broken. (look at flash animation on hydrolysis)

Chemistry

2 answers:

Dvinal [7]3 years ago

7 0

Water must be added to break the bond

ella [17]3 years ago

4 0

The correct answer is water.

During hydrolysis, _____water______ must be added before the bonds can be broken.

For example, sucrose undergoes hydrolysis to break into glucose and fructose. Here sucrose is a disaccharide and glucose and fructose are monosaccharides.

Sucrose + H₂O → glucose + fructose

Esters are formed by the combination of carboxylic acid and alcohol.

But the hydrolysis of ester causes the release of carboxylic acid and alcohol.

RCOOR'(ester) + H₂O → RCOOH(carboxylic acid) + R'OH(alcohol)

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A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal

devlian [24]

Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

6 0

3 years ago

A mixture of CS2(g) and excess O2(g) is placed in a 10 L reaction vessel at 100.0 ∘C and a pressure of 3.10 atm . A spark causes

ziro4ka [17]

Answer:

PCO2 = 0.6 25 atm

PSO2 = 1.2 75 atm

PO2 = 0.6 atm

Explanation:

Step 1: Data given

Volume = 10.0 L

Temperature = 100.0 °C

Pressure = 3.10 °C

After reaction, the temperature returns to 100.0 ∘C, and the mixture of product gases (CO2, SO2, and unreacted O2) is found to have a pressure of 2.50 atm

Step 2: The balanced equation

CS2(g)+3O2(g)→CO2(g)+2SO2(g)

Step 3: Name the reactants and products

a = CS2

b = O2 before reaction

c = CO2

d = SO2

e = nS O2 after reaction with n = the number of moles

Step 4: Calculate moles before reaction

PV = nRT

n = PV/(RT)

(na + nb) = (3.10atm) * (10.0L) / ((0.08206 Latm/moleK) * (373.15K))

(na + nb) = 1.0124

Step 5: Calculate moles after reaction

PV = nRT

n = PV/(RT)

nc + nd + ne) = PV/(RT) = (2.50 atm)*(10.0L) / ((0.08206 Latm/moleK)*(373.15K))

(nc + nd + ne) = 0.816 moles

Step 6: Calculate mol fraction

For 1 mole CS2 we need 3 moles O2 to produce 1 mole of CO2 and 2 moles of SO2

moles O2 remaining = ne = nb - 3na

moles CO2 produced = nc = na

moles SO2 producted = nd = 2na

(nc + nd + ne) = 0.816 moles = nb - 3na + na + 2na = 0.816

nb = 0.816

. (na + nb) = 1.0124

na = 1.0124 moles - 0.816 moles = 0.208

which leads to

nc = na = 0.208

nd = 2na = 2*0.208 = 0.416

ne = 0.816 - 3*0.208 = 0.192

mole fraction CO2 = 0.208 / (0.208 + 0.416 + 0.192) = 0.25

mole fraction SO2 = 0.416 / (0.208 + 0.416 + 0.192) = 0.5 1

mole fraction O2 = 0.192 /(0.208 + 0.416 + 0.192) = 0.24

Step 6: Calculate partial pressure

PCO2 = 0.25 * 2.50 atm = 0.6 25 atm

PSO2 = 0.51 * 2.50 atm = 1.2 75 atm

PO2 = 0.24 * 2.50 atm = 0.6 atm

Step 7: Control results

now let's verify a couple of things

PV = nRT

P = nRT/V

before rxn

P = (0.208 + 0.816) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 3.10 atm

after rxn

P = ((0.208 +0.416+0.192) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 2.50 atm

8 0

2 years ago

How many atoms are in a molecule of hydrogen peroxide?

Alenkasestr [34]

2 Hydrogen and 2 Oxygen

7 0

3 years ago

During an investigation, similar glow sticks were placed in two beakers containing water at different temperatures. A record of

allochka39001 [22]

Answer:

i got you dawg just gimme one sec i'll get to you fr g

Explanation:

6 0

2 years ago

Read 2 more answers

What design element of an astronaut's spacesuits protects him or her from micrometeoroids, which are tiny bits of space debris t

leonid [27]

Answer:

Tough outer shell

Explanation:

Three types of space suits based on their purpose are IVA (Intravehicular activity), EVA (extravehicular activity), and IEVA (intra/extravehicular activity)

Space suits have an outer layer known as the Integrated Thermal Micrometeoroid Garment also known as the TMG or ITMG

The functions of the TMG includes the provision of an insulation medium to the wearer of the suit thereby preventing loss of heat and keeping the wearer warm, protect the wearer from solar radiation which are harmful as well as from Micrometeoroid and debris orbiting in space known as MMOD

The outermost layer of the EMU TMG suits is white Ortho-Fabric that consists of Nomex, Kevlar and Gore-Tex, making it fire resistant, very strong and very tough, and water proof as well as breathable.

3 0

3 years ago