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tigry1 [53]
2 years ago
8

The law of the conservation of mass must be obeyed in a chemical reaction. what do the atoms do during a reaction that shows the

y obey the conservation of mass?
Chemistry
1 answer:
Rainbow [258]2 years ago
8 0
A balanced equation is a prime example of the law of the conservation of mass as the number of atoms in the reactants is consistent with the number of atoms in the reactants meaning the amount of matter has not changed and no mass has been created or destroyed hence obeying the law.
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A cloud type that usually produces precipitation is a(n)
abruzzese [7]
The cloud that produces rain is the cumulonimbus cloud.
5 0
3 years ago
A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is
Airida [17]

Answer:

the maximum safe operating temperature the engineer should recommend for this reaction is 616 °C  

Explanation:

Given the data in the question;

First we calculate the Volume of the steel cylinder;

V = πr²h

radius r = Diameter / 2 = 27 cm / 2 = 13.5 cm

height h = 32.4 cm

so we substitute

V = π × ( 13.5 cm )² × 32.4 cm

V  = π × 182.25 cm × 32.4 cm

V = 18550.79 cm³  

V = 18.551 L

given that; maximum safe pressure P = 3.10 MPa = 30.5946 atm

vessel contains 0.218kg or 218 gram of carbon monoxide gas

molar mass of carbon monoxide gas is 28.010 g/mol

so

moles of carbon monoxide gas n = 218 gram /  28.010 g/mol = 7.7829 mol

we know that;

PV = nRT

solve for T

T = PV / nR

we know that gas constant R = 0.0820574 L•atm•mol⁻¹ K⁻¹

so we substitute

T = ( 30.5946 × 18.551 ) / ( 7.7829 × 0.082 )

T = 567.5604 / 0.6381978

T = 889.317387 K

T = ( 889.317387 - 273.15 ) °C

T = 616.167 ≈ 616 °C  { 3 significant digits }

Therefore, the maximum safe operating temperature the engineer should recommend for this reaction is 616 °C  

6 0
2 years ago
Please help, I really don’t understand this!!!
kap26 [50]

<u>Analysing the Question:</u>

We are given the balanced equation:

C₆H₁₂O₆ + 6O₂→ 6CO₂ + 6H₂O

from this equation, we can say that: <em>for every 1 mole of Glucose, we need 6 moles of Oxygen</em>

<u>Moles of Glucose used in the reaction:</u>

Molar mass of Glucose = 180 grams / mol

Given mass of Glucose = 1 gram

Mole of Glucose = Given mass / Molar mass

Moles of Glucose = 1 / 180 moles

<u>Mass of Oxygen required:</u>

We know that for every mole of Glucose, we need 6 moles of Oxygen

So, for 1/180 moles of Glucose, we need 6 / 180 = 1 / 30 moles of Oxygen

Mass of 1 / 30 moles of Oxygen:

Mass = Molar mass * number of moles

Mass of Oxygen = 32 * 1/30

Mass of Oxygen  = 32 / 30

Mass of Oxygen = 1.06 grams

5 0
3 years ago
Determine the oxidation state of each of the following species.<br> Pb2+
mafiozo [28]

Answer:

Pb oxidation number is +2

C in CH4 is -4. H is +1. 4H + 1C = 0. ; 4(+1) + C = 0 ; C = -4

O is usually -2. 3O = 3(-2) = -6. ; 2Fe =. +6 ; each Fe is +3

2Ag = +2 since O = -2 ; each Ag = +1

6 0
3 years ago
g Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The h
Setler79 [48]

Answer:

6.88 mg

Explanation:

Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄

The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.

175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P

Step 2: Calculate the rate constant for the decay of ³²P

The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.

k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹

Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days

For first-order kinetics, we will use the following expression.

ln P = ln P₀ - k × t

ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d

P = 6.88 mg

3 0
3 years ago
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