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solmaris [256]
3 years ago
7

Standard 6 multiple choice questions English paper​

Chemistry
2 answers:
Charra [1.4K]3 years ago
5 0
So what’s the question? I’m confused
nikdorinn [45]3 years ago
5 0

Answer:

please attach the images in the attachment and then I 'll b able to answer your questions regarding any subject!!

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2-An analysis that establishes the chemical identity of the species
saw5 [17]

Answer:

C Qualitative Analysis.

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3 years ago
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Your observation and assessment data should help you _______ your curriculum A. change B. modify and individualize C. evaluate t
joja [24]
Hello,
I believe the answer is C. A and B say that it would help to change it but when we think about it data and opinions are not what people use to change a curriculum or program. Mainly they use surveys, which they didn't use here. Option D doesn't make any sense, data doesn't complement a program it supports some sort of theory which leads us to C. C is the only option where they use the data for effectiveness. Hope this helps!
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3 years ago
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What happens to the electrons in an ionic bond?
Anton [14]

Answer:

In an ionic bonds, the metal loses electrons to become a positively charged cation, In which the nonmetal accepts those electrons to become a negatively charged anion.

Explanation:

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3 years ago
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A class wants to find the ideal temperature for plants to grow in a terrarium. If the students set up several terrariums, what s
Veseljchak [2.6K]

It would be B because you want to know which TEMPERATURE is the best for the plant. Different brightness for a bulb will give different amounts of heat. Since the amount of heat is your INDEPENDENT VARIABLE, you have to change the amount of heat for each plant, so the brightness of bulbs. Therefore, B is your answer!

3 0
4 years ago
How many days does it take 16.Og of Gold-198 to decay to 1.0g? (each half-
forsale [732]

Answer:

10.8 days (3 sig.figs.)

Explanation:

All radioactive decay is 1st order decay defined by the expression A = A₀e^-kt

which is solved for time of decay (t) => t = ln(A/A₀) / -k

A = final weight = 1.0 gram

A₀ = initial weight = 16.0 grams

k = rate constant = 0.693/t(1/2) = 0.693/2.69 days = 0.258 days⁻¹

t = ln(1/16) / -0.258da⁻¹ = (-2.77/-0.258) days = 10.74646792 days (calculator)

≅ 10 days (1 sig. fig. based on given 1 gram mass)

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3 years ago
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