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kramer
3 years ago
11

Plz help all chemist!

Chemistry
1 answer:
solmaris [256]3 years ago
6 0

Answer:

D. bromine

Explanation:

Highest electronegativity in the period has element closest to the *A group

K Ca Cu Br

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The mixture of carbon dioxide and water tastes salt why​
vampirchik [111]

Answer:

When carbon dioxide reacts with water carbonic acid is formed, from which hydrogen ions dissociate increasing the acidity of the systemCarbon dioxide emissions to the atmosphere can therefore increase the acidity of land, sea and air

Explanation:

6 0
3 years ago
A sample of O2 gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of CO2 to effuse
ludmilkaskok [199]

Answer:

\large \boxed{\text{5.9 s}}

Explanation:

Graham’s Law applies to the effusion of gases:

The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

r \propto \dfrac{1}{\sqrt{M}}

If you have two gases, the ratio of their rates of effusion is

\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}

The time for diffusion is inversely proportional to the rate.

\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}

Let CO₂ be Gas 1 and O₂ be Gas 2

Data:

M₁ = 44.01

M₂ = 32.00

Calculation

\begin{array}{rcl}\dfrac{t_{2}}{t_{1}} & = & \sqrt{\dfrac{M_{2}}{M_{1}}}\\\\\dfrac{t_{2}}{\text{5 s}}& = & \sqrt{\dfrac{44.01}{32.00}}\\\\& = & \sqrt{1.375}\\t_{2}& = & \text{5 s}\times 1.173\\& = & \mathbf{5.9 s} \\\end{array}\\\text{It will take $\large \boxed{\textbf{5.9 s}}$ for the carbon dioxide to effuse.}

4 0
3 years ago
Materials are transported within a single celled organism by the
CaHeK987 [17]
The cytoplasm transports materials within a single celled organism. that is the answer.
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True or False. Star gazing is a pretty popular hobby in today's world.
ValentinkaMS [17]

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yes (true)

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Star gazing has started to become a more popular hobby in todays world due to the pandemic

7 0
3 years ago
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Convert 0.74Kcal/min to cal/sec
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<span>0.74 Kcal/min x 1000 cal/60 s         


</span>
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