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2 years ago

How do you do problem #15?

Chemistry

1 answer:

SVETLANKA909090 [29]2 years ago

7 0

You would convert the cg to g. 1.66 cg is 0.0166 g. then you add. 0.398 + 0.0166 = 0.4146. the answer is 0.4146 grams.

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You have a 25.2 L sample of gas at 1.25 atm and 25.0 degrees Celsius. How many moles are present in this gas. For your answer, p

Elenna [48]

Answer:

<u>1.29 mol</u>

Explanation:

This is a direct application of the equation for ideal gases.

$PV=nRT$

Where:

P = pressure = 1.25 atm
V = volume = 25.2 liter
R = Universal constant of gases = 0.08206 atm-liter/K-mol
T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
n = number of moles

Solving for n:

$n=\frac{PV}{RT}$

Substituting:

$n=\frac{1.25atm\times 25.2liter}{0.08206atm-liter/K-mol\times298.15K }\\\\n=1.29mol$

8 0

3 years ago

Which two chemical substances are needed to begin the photosynthesis factory process?

Rudiy27

<span>Carbon dioxide CO2 and water H2O. Through photosynthesis makes sugar C6H12O6.</span>

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2 years ago

Which of the following is required to change the state of matter?

nignag [31]

Since there are no given items, I will give a general answer. Energy....or the lack of it. Examples: Heat, electricity, force (when an item is moving and it impacts something, it heats up...friction is an example of this), etc

3 0

3 years ago

Read 2 more answers

How does botulinum toxin block synaptic transmission?

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It blocks ACh release so the muscle cannot contract

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3 years ago

Citric acid is a naturally occurring compound. what orbitals are used to form each indicated bond? be sure to answer all parts.2

Rina8888 [55]

Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.

Answer 1) σ Bond a : C has $SP^{2}$ O has $SP^{2}$ .

Explanation : The orbitals of oxygen and carbon which are involved in $SP^{2}$ hybridization to form sigma bonds. This is observed at 'a' position in the citric acid molecule.

Answer 2) π Bond a: C has π orbitals and O also has π orbitals.

Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.

Answer 3) Bond b: O $SP^{3}$ H has only S orbital involved in bonding.

Explanation : The bonding at 'b' position involves oxygen $SP^{3}$ hydrogen atoms in it. It has $SP^{3}$ hybridized orbitals and S orbital of hydrogen involved in the bonding.

Answer 4) Bond c: C is $SP^{3}$ O is also $SP^{3}$

Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of $SP^{3}$ hybridized bonds.

Answer 5) Bond d: C atom has $SP^{3}$ C atom has $SP^{3}$

Explanation : At the position of 'd' the bonding between two carbon atoms is found to be $SP^{3}$ . Therefore, the orbitals that undergo $SP^{3}$ hybridization are $SP^{3}$ .

Answer 6) Bond e : C1 containing O $SP^{2}$

C2 is $SP^{3}$

Explanation : The carbon atom which contains oxygen along with a double bond has $SP^{2}$ hybridized orbitals involved in the bonding process; whereas the carbon at C2 has $SP^{3}$ hybridized orbitals involved during the bonding. This is for the 'e' position.

7 0

3 years ago