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Alex Ar [27]
3 years ago
13

Clouds are named for 3 different attributes including shape, _____________ & precipitation. *

Chemistry
2 answers:
mart [117]3 years ago
4 0

Answer:

color is there right answer for this question please make me a brainliest and follow me please please please please

Burka [1]3 years ago
3 0
Shape, COLOR, precipitation
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What would cause the pressure of a gas to increase?​
IrinaK [193]

Answer: The pressure of a gas will increase when there is a decrease in the volume of the gas.

Explanation: according to Boyle's law, the volume of a gas will decrease when the pressure is increased at constant temperature and vice versa.

8 0
3 years ago
One brand of vinegar has a pH of 4.5. Another brand has a pH of 5.0. The equation for the pH of a substance is pH = –log[H+], wh
grigory [225]
[H+] in first brand:
4.5 = -log([H+])
[H+] = 10^(-4.5)

[H+] in second brand:
5 = -log[H+]
[H+] = 10^(-5)

Difference = 10^(-4.5) - 10^(-5)
= 2.2 x 10⁻⁵

The answer is A.
7 0
3 years ago
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How does stem apply to science?<br><br> (i swear if yall just answer for points ima lose it)
vova2212 [387]

STEM is an acronym meaning

Science

Technology

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8 0
3 years ago
Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature
weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}     ------>NO_{(g)}

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

\delta H^0__{R,T_2} = \delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'

where:

\delta H^0__{R} = enthalpy of reaction

{\delta C_p(T')} = the difference in the heat capacities of the products and the reactants.

∴

\delta H^0__{R,435K} = \delta H^0__{R,298.15K} + \int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'

= 1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

\delta H^0__{R,435K} ≅ 91383 J

6 0
3 years ago
Balance C12H22011 → C + H2
svlad2 [7]

Answer:

What is this?

Explanation:

7 0
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