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Allushta [10]
3 years ago
5

PLEASE SOMEBODY HELP PLEASE ! I WILL MARK BRAINLY I PROMISE !

Chemistry
1 answer:
IgorLugansk [536]3 years ago
5 0
A. your reactants are on the left of the arrow.
B. your products are on the left of the arrow.
C.its read as a yeild sign.
D.the s means it's a soild and the g means it's a gas.
You might be interested in
What electrically neutral atom has 30 neutrons and 25 electrons?
Black_prince [1.1K]

Answer:

Manganese (Mn)

Explanation:

We know it's manganese because we are told it is an electrically neutral atom. This means it has the same number of protons and electrons. If it has 25 electrons, it has 25 protons. Protons tell us the atomic number of the atom, which also tells us the name of the element. Manganese is element 25 on the periodic table.

7 0
3 years ago
If you have a 4.6 L of gas in a piston at a pressure of 1.2 atm and compress 2p the gas until its volume is 2.6 L, what will the
Nesterboy [21]

Answer:

2.12atm

Explanation:

Boyle's Law: P1V1 = P2V2

Manipulate to solve for unknown: P2 = P1V1/V2

Substitute values:  P2=(1.2atm)(4.6L)/2.6L

P2 = 2.1230769atm

Round to 3 sig figs to get 2.12atm

3 0
3 years ago
What is the pH of 0.30 M ethanolamine, HOCH2CH2NH2, (Kb = 3.2 x 10−5)?
Dmitry [639]

Answer:

pH= 11.49

Explanation:

Ethanolamine is an organic chemical compound of the formula; HOCH2CH2NH2. Ethanolamine, HOCH2CH2NH2 is a weak base.

From the question, the parameters given are; the concentration of ethanolamine which is = 0.30M, pH value= ??, pOH value= ??, kb=3.2 ×10^-5

Using the formula below;

[OH^-]=√(kb×molarity)----------------------------------------------------------------------------------------------------------(1)

[OH^-] =√(3.2×10^-5 × 0.30M)

[OH^-]= √(9.6×10^-6)

[OH^-]=3.0984×10^-3

pOH= -log[OH^-]

pOH= -log 3.1×10^-3

pOH= 3-log 3.1

pH= 14-pOH

pH= 14-(3-log3.1)

pH= 11+log 3.1

pH= 11+ 0.4914

pH= 11.49

8 0
3 years ago
Count the total number of atoms in sio2:
grigory [225]
Atoms are the basic units of matter and the defining structure of elements. We count the number of atoms by the total number of elements present in the compound. In this case, we have 1 atom of Si and 2 atoms of oxygen which would have 3 total number of atoms.
8 0
3 years ago
Use standard enthalpies of formation to calculate Δ H ∘ rxn for each reaction. MISSED THIS? Read Section 7.9; Watch KCV 7.9, IWE
Eva8 [605]

Answer:

Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.

C₂H₄ (g), 52.5 kJ/mol

H₂ (g), 0 kJ/mol

C₂H₆ (g), -83.7 kJ/mol

CO (g), -110.525 kJ/mol

H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

HNO₃ (aq), -206.28 kJ/mol

NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol

ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol

ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol

ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol

ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×-206.28) + (1×90.29) = -322.27 kJ/mol

ΔH∘(reactants) = (3×33.2) + (1×-285.8) = -186.2 kJ/mol

ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol

ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol

ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol

Hope this Helps!!!

4 0
3 years ago
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