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2 years ago

A scientist obtains a set of measurements that are very close to one another. These can be said to be:

Chemistry

1 answer:

pickupchik [31]2 years ago

3 0

Answer:

I believe this is a K-12 test question. If the answers below are what you have on your test . . .

- Precise

- Accurate

- Identical

- None of the above

Then the answer is <u>precise</u>.

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Which of these is a chemical property of a substance? texture ductility reactivity conductivity

beks73 [17]

Reactivity is a chemical property of a substance. According to EPA regulations, it is normally unstable and readily

undergoes violent change without detonating. it can explode or violently react when exposed to water, when heated, or under STP.

8 0

3 years ago

Read 2 more answers

Can someone answer this question? It's not that long

Kitty [74]

Answer:

sections 2 and 3

Explanation:

logic

6 0

2 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago

As water molecules move through the hydrological cycle, water both gains and loses energy. In which case would the water be losi

Triss [41]

Condensation. I just completed the test

5 0

3 years ago

Read 2 more answers

The radioactive decay of a certain sample produced 846 disintegrations per minute. exactly 3.00 days later, the rate of decay wa

Margarita [4]

Answer:

$\boxed{\text{1.81 da}}$

Explanation:

1. Calculate the decay constant

The integrated rate law for radioactive decay is 1

$\ln\dfrac{A_{0}}{A_{t}} = kt$

where

A₀ and A_t are the counts at t = 0 and t

k is the radioactive decay constant

$\ln \dfrac{846}{269} = k \times 3.00\\\\\ln3.145 = 3.00k\\1.146 = 3.00k\\\\k =\dfrac{1.146}{3}\\\\k = \text{0.382 /da}\\$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{0.382} = \text{1.81 da}$

The half-life for decay is $\boxed{\textbf{1.81 da}}$ .

3 0

3 years ago