Very easy significant figures question!

The alkali earth metal beryllium (Be) engages in a chemical reaction and loses all of its valence electrons.

denpristay [2]

The loss of electron from an results in the formation of cation represented by the positive charge on the element whereas gaining of electron results in the formation of anion represented by the negative charge on the element.

The alkali earth metal beryllium ( $Be$ ) belongs to the second group of the periodic table. The ground state electronic configuration of $Be$ is: $1s^{2}2s^{2}$

From the electronic configuration it is clear that it has 2 valence electrons in its valence shell ( $2s^{2}$ ).

After losing all valence electrons that is 2 electrons from $2s$ orbital. The electronic configuration will be:

$1s^{2}2s^{0}$

Since, lose of electron is represented by positive charge on the element symbol. So, the beryllium will have +2 charge on its symbol as $Be^{2+}$ .

Hence, beryllium will have 2+ charge on it after losing all its valence electrons in the chemical reaction.

6 0

3 years ago

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Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

8 0

3 years ago

Which of the following is the best example of an open system? (2 points)

Dmitriy789 [7]

1) A Car engine.............. Hope it helps, Have a nice day :)

5 0

3 years ago

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What is the temperature of 0.645 mole of neon in a 2.00 L vessel at 4.68 atm?

storchak [24]

Answer:

—96.03°C

Explanation:

We'll begin by writing out the information provided by the question. This includes:

Number of mole (n) = 0.645 mole

Volume (V) = 2.00 L

Pressure (P) = 4.68 atm

Temperature (T) =?

Recall: that the gas constant = 0.082atm.L/Kmol

With the ideal gas equation PV = nRT, the temperature of the gas can be obtained as follow:

PV = nRT

4.68 x 2 = 0.645 x 0.082 x T

Divide both side 0.645 x 0.082

T = (4.68 x 2) /(0.645 x 0.082)

T = 176.97 K

Now, We can also express the temperature obtained in celsius as shown below:

Temperature (celsius) = temperature (Kelvin) - 273

Temperature (celsius) = 176.97 - 273

Temperature (celsius) = —96.03°C

The temperature of the Neon gas is

—96.03°C

7 0

3 years ago

What processes in the water cycle takes water from oceans and land masses?

krek1111 [17]

Answer:

Water Cycle

Earth is a truly unique in its abundance of water. Water is necessary to sustaining life on Earth, and helps tie together the Earth's lands, oceans, and atmosphere into an integrated system. Precipitation, evaporation, freezing and melting and condensation are all part of the hydrological cycle - a never-ending global process of water circulation from clouds to land, to the ocean, and back to the clouds.
This cycling of water is intimately linked with energy exchanges among the atmosphere, ocean, and land that determine the Earth's climate and cause much of natural climate variability.
The impacts of climate change and variability on the quality of human life occur primarily through changes in the water cycle. As stated in the National Research Council's report on Research Pathways for the Next Decade (NRC, 1999): "Water is at the heart of both the causes and effects of climate change."

<h2>Importance of the ocean in the water cycle</h2>

The ocean plays a key role in this vital cycle of water.
The ocean holds 97% of the total water on the planet; 78% of global precipitation occurs over the ocean, and it is the source of 86% of global evaporation.
Besides affecting the amount of atmospheric water vapor and hence rainfall, evaporation from the sea surface is important in the movement of heat in the climate system.
Water evaporates from the surface of the ocean, mostly in warm, cloud-free subtropical seas.
This cools the surface of the ocean, and the large amount of heat absorbed the ocean partially buffers the greenhouse effect from increasing carbon dioxide and other gases.
Water vapor carried by the atmosphere condenses as clouds and falls as rain, mostly in the ITCZ, far from where it evaporated, Condensing water vapor releases latent heat and this drives much of the the atmospheric circulation in the tropics.
This latent heat release is an important part of the Earth’s heat balance, and it couples the planet’s energy and water cycles.

The major physical components of the global water cycle include the evaporation from the ocean and land surfaces, the transport of water vapor by the atmosphere, precipitation onto the ocean and land surfaces, the net atmospheric transport of water from land areas to ocean, and the return flow of fresh water from the land back into the ocean.
. The additional components of oceanic water transport are few, including the mixing of fresh water through the oceanic boundary layer, transport by ocean currents, and sea ice processes.
On land the situation is considerably more complex, and includes the deposition of rain and snow on land; water flow in runoff; infiltration of water into the soil and groundwater; storage of water in soil, lakes and streams, and groundwater; polar and glacial ice; and use of water in vegetation and human activities.
Illustration of the water cycle showing the ocean, land, mountains, and rivers returning to the ocean.
Processes labeled include: precipitation, condensation, evaporation, evaportranspiration (from tree into atmosphere), radiative exchange, surface runoff, ground water and stream flow, infiltration, percolation and soil.

6 0

2 years ago