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wariber [46]
4 years ago
10

Solution A has a pH of 9, while solution B has a pH of 8. How much more basic is solution A than

Chemistry
1 answer:
klio [65]4 years ago
7 0

Answer:

I'm pretty sure the correct answer is b, I took this quite a while ago so I might not be right I would do some research about it. hope I helped!

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“The element vanadium is a solid at room temperature, and its density is 6110 kilograms per meter cubed. Its melting point is 19
Luba_88 [7]
I think the correct answer is the second option. The statement pertaining to the properties of the element vanadium is reliable because the facts presented are testable or can be verified by doing experimentation. Also, nowadays, many studies are present that contains properties of almost anything.
7 0
3 years ago
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Given the following balanced equation, determine the rate of reaction with respect to [Cl2]. If the rate of Cl2 loss is 4.44 × 1
PolarNik [594]

Answer:

8.88 x 10⁻² M/s

Explanation:

The rate  of reaction for:

                    NO(g) + Cl₂ (g) ⇒  2NOCl(g)

is rate = -ΔNO/Δt = -ΔCl2/Δt = 1/2 ΔNOCl/Δt

so  ΔNOCl/Δt = 2 ΔCl2/Δt  = 2  x 4.44 × 10⁻²  M/s = 8.88 x 10⁻² M/s

In general given a reaction

                            aA + bB ⇒ cC + dD

rate = -1/a ΔA/Δt = -1/b ΔB/Δt = 1/c ΔC/Δt = 1/d ΔD/Δt

8 0
3 years ago
which equation is set up correctly to determine the volume of a 1.5 mole sample of oxygen gas at 22 C and 100kPa
Murljashka [212]
Under STP condition, the gas has a rule of 22.4 L per mole. And according to the ideal gas law, V1/T1=V2/T2. Under STP, 1.5 mol gas has volume of 33.6 L. So the volume under 22 C is 33.6*295/273=36.3 L.
7 0
4 years ago
Please help its due today..!
gogolik [260]

Answer:

This all should be correct:

c

a

c

a

6 0
4 years ago
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For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.
Rudik [331]

Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

8 0
3 years ago
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