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Leto [7]
4 years ago
11

Which statement is not true? *

Chemistry
1 answer:
Ronch [10]4 years ago
7 0

bases have a ph lower than 7 is false. it's actually acids that have a ph lower than 7, 7 is neutral, and above 7 to 14 is basic.

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A battery
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2. What is one effect of convection currents in the mantle?
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D no. is the answer of your question

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3. If 175 grams of nitrogen reacted with hydrogen to make hydrogen, how much ammonia would be needed?
zmey [24]

Answer:

37.5 g of H₂

Explanation:

N₂ (28g) + 3H₂ (6g) => 2NH₃ (34g)

Every 28g of N₂ needs 6g of H₂

=> Every 7g of N₂ needs 1.5g of H₂

=> 175g of N₂ needs 1.5 x 25 g of H₂

=> 37.5 g of H₂

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2 years ago
Read the passage about the fluoridation of drinking water. Since the early 1960s, fluoride has been added to many sources of dri
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Assignment

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It is difficult to tell if fluoride causes this cancer.

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6 0
4 years ago
Alice and Bob are experimenting with two moles of neon, a monatomic gas, that starts out at conditions of standard temperature a
Archy [21]

Answer:

29273.178 joules have been added to the gas for the entire process.

Explanation:

The specific heats of monoatomic gases, measured in joules per mol-Kelvin, are represented by the following expressions:

Isochoric (Constant volume)

c_{v} = \frac{3}{2}\cdot R_{u} (1)

Isobaric (Constant pressure)

c_{p} = \frac{5}{2}\cdot R_{u} (2)

Where R_{u} is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.

Under the assumption of ideal gas, we notice the following relationships:

1) Temperature is directly proportional to pressure.

2) Temperature is directly proportional to volume.

Now we proceed to find all required temperatures below:

(i) <em>Alice heats the gas at constant volume until its pressure is doubled</em>:

\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}} (3)

(\frac{P_{2}}{P_{1}} = 2, T_{1} = 273.15\,K)

T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}

T_{2} = 2\times 273.15\,K

T_{2} = 546.3\,K

(ii) <em>Bob further heats the gas at constant pressure until its volume is doubled</em>:

\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}} (4)

(\frac{V_{3}}{V_{2}} = 2, T_{2} = 546.3\,K)

T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}

T_{3} = 2\times 546.3\,K

T_{3} = 1092.6\,K

Finally, the heat added to the gas (Q), measured in joules, for the entire process is:

Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})] (5)

If we know that R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}, n = 2\,mol, T_{1} = 273.15\,K, T_{2} = 546.3\,K and T_{3} = 1092.6\,K, the heat added to the gas for the entire process is:

c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)

c_{v} = 12.471\,\frac{J}{mol\cdot K}

c_{p} = \frac{5}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)

c_{p} = 20.785\,\frac{J}{mol\cdot K}

Q = (2\,mol)\cdot \left[\left(12.471\,\frac{J}{mol\cdot K} \right)\cdot (536.3\,K-273.15\,K)+\left(20.785\,\frac{J}{mol\cdot K} \right)\cdot (1092.6\,K-546.3\,K )\right]

Q = 29273.178\,J

29273.178 joules have been added to the gas for the entire process.

6 0
3 years ago
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