The nucleus of the atom contains what subatomic particle(s)? Protons only Neutrons only Electrons only Protons and neutrons Protons and electrons

Answer:

water carbon dioxide

Explanation:

Answer is: the objects mechanical energy is 52,92 J.
m(object) = 0,3 kg.
h(object) = 18 m.
g = 9,8 m/s².
E(object) = m·g·h.
E(object) = 0,3 kg · 9,8 m/s² · 18 m.
E(object) = 52,92 N·m = 52,92 J..
g - the acceleration of free fall.
mg - weight of the object.

Answer :

Part A : The value of [tex]\alpha, \beta, \gamma \text{ and }\delta[/tex] are 2, 19, 12 and 14 respectively.

Part B : [tex]6.0\times 10^1moles[/tex] of [tex]O_2[/tex] are required to react with 6.4 moles of [tex]C_6H_{14}[/tex].

Solution :

Part A :

The given unbalanced equation is,

[tex]\alpha C_6H_{14}(g)+\beta O_2(g)\rightarrow \gamma CO_2(g)+\delta H_2O(g)[/tex]

The balanced equation is,

[tex]2C_6H_{14}(g)+19O_2(g)\rightarrow 12CO_2(g)+14H_2O(g)[/tex]

Part B :

From the balanced equation, we conclude that

2 moles of [tex]C_6H_{14}[/tex] react with 19 moles of [tex]O_2[/tex]

6.4 moles of [tex]C_6H_{14}[/tex] react with [tex]\frac{19moles}{2moles}\times 6.4moles=60.8moles=6.0\times 10^1moles[/tex] of [tex]O_2[/tex]

Therefore, [tex]6.0\times 10^1moles[/tex] of [tex]O_2[/tex] are required to react with 6.4 moles of [tex]C_6H_{14}[/tex].

The balanced equation is 2C3H7(g) + 9O2(g) → 6CO2(g) + 7H2O(g), therefore the coefficients α, β, γ, δ are 2,9,6,7, respectively. And 29 moles of O2 are required to react completely with 6.4 moles of C6H14.

For part a, to balance the equation for the combustion of hexane, we have to adjust the coefficients in the equation C6H14(g) + αO2(g) → βCO2(g) + γH2O(g) such that the number of atoms of each element on both sides of the equation are equal. The balanced equation becomes: 2C3H7(g) + 9O2(g) → 6CO2(g) + 7H2O(g) . It means that α, β, γ, δ = 2, 9, 6, 7 respectively.

For part b, according to the balanced equation, 9 moles of O2 are required to react completely with 2 moles of C6H14. So, if there are 6.4 moles of C6H14, you will need: (6.4 moles C6H14 * 9 moles O2) / 2 moles C6H14 = 28.8 ≈ 29 moles of O2. In this case  'n' equals 29 moles.

https://brainly.com/question/28294176

#SPJ3

The conjugate acid of HPO2- is H2PO4-.

The conjugate acid of HPO2- is H2PO4-.

Final answer:

To determine the equilibrium pressure of NO, set up an ICE table using the values of the equilibrium constant (Kc) and initial partial pressures. Assume the increase in NO pressure is 'x' and the decrease in N2O and O2 pressures is '2x' and 'x', respectively. Calculate the equilibrium pressures by adding the initial pressures to their corresponding changes.

Explanation:

To determine the equilibrium pressure of NO, we need to set up an ICE table and use the values of the equilibrium constant (Kc) and initial partial pressures of the three gases. Let's assume the increase in the pressure of NO is 'x'. Since the stoichiometric coefficient of NO is 2, the decrease in the pressures of N2O and O2 will be '2x' and 'x', respectively. The equilibrium partial pressures of the three gases can be calculated by adding the initial pressures to their corresponding changes.

Initial pressures:

Change in pressures:

Equilibrium pressures:

Since the equilibrium constant (Kc) is given as 1.71 × 10-1, which represents the ratio of the equilibrium concentrations of products to reactants, we can set up the following equation:

Kc = [NO]2 / ([N2O] * [O2])

Substituting the equilibrium pressures into this equation:

1.71 × 10-1 = ([1.96 × 10-3 + x]2 / ([1.96 × 10-3 - 2x] * [1.96 × 10-3 - x]))

Now, solve this equation to find the value of 'x', which represents the equilibrium increase in the pressure of NO.

Once 'x' is determined, substitute it back into the equations for the equilibrium pressures to calculate the final equilibrium pressures of N2O, O2, and NO.

Ocean exploration can lead to cultural and historical insights through the excavation of shipwrecks, aid in the discovery and mapping of natural resources like oil, and contribute to marine biology by discovering new species, ecosystems, and fossil sites.

A possible application for discoveries made during ocean exploration includes the excavation of historical shipwrecks to shed light on cultural interactions and trade routes of the past. For instance, the excavation of the Byzantine Serçe Limanı Shipwreck provides valuable information about cross-cultural artistic exchanges during the Middle Byzantine period. Similarly, studies of the Battle of Bạch Đằng help map the cultural landscape associated with historical Vietnamese maritime events. Ocean exploration can also lead to the discovery and mapping of natural resources such as oil deposits, which is vital for their utilization, drawing a parallel to the need for exploration before the extraction of resources, much like we can't produce more oil than we discover. Additionally, marine ecologists involved in ocean exploration contribute to the study of marine life and ecosystems, which includes discovering new fossil sites, organizing and classifying organisms, and identifying new species.

Ocean exploration can lead to potential ocean floor colonization, historical insights from shipwrecks, new resource discoveries, and innovations in resource extraction methods. Marine ecologists also benefit from discoveries made, which contribute to scientific and environmental understanding.

Discoveries made during ocean exploration have a multitude of applications, one of which is the possibility of colonizing the ocean floor. This could serve as a precursor to space colonization due to similar challenges faced in both environments, such as the development of sealed-off, self-sustaining habitats. Additionally, research on historic shipwrecks, like the Byzantine Serçe Limanı Shipwreck, provides invaluable insights into historical trade routes and cross-cultural interactions.

Further benefits of ocean exploration include advancements in navigational technology, which were historically influenced by the contributions of Muslim and Chinese inventions. These innovations enabled European explorers to sail across the Atlantic Ocean, leading to the discovery of new trade routes and the establishment of new markets. Ocean exploration also plays a critical role in uncovering new natural resources, like oil deposits, and in the development of innovative methods to extract and utilize these resources effectively.

Moreover, a marine ecologist might conduct studies to discover new fossil sites, classify organisms, or name new species, underlining the scientific and environmental significance of oceanic discoveries.

Final answer:

The theoretical yield of nitric acid, when 2.0 moles of nitrogen dioxide react with sufficient oxygen, is calculated using stoichiometry and comes out to be 126.02 grams.

Explanation:

The question involves using a chemical equation and stoichiometry to calculate the theoretical yield of a compound, specifically nitric acid (HNO3), from a given amount of reactants. We start by balancing the chemical equation: 4NO2(g) + O2(g) + 2H2O(l) = 4HNO3(aq). This shows that 4 moles of NO2 react with 1 mole of O2 to produce 4 moles of HNO3. With 2.0 moles of NO2 present, NO2 is the limiting reactant since it reacts in a 4:1 ratio with O2, and there is enough O2 provided (2.0 moles).

To calculate the theoretical yield of nitric acid, we use the molar ratio from the balanced equation. Since 4 moles of NO2 produce 4 moles of HNO3, 2.0 moles of NO2 will produce 2.0 moles of HNO3. Now, we need to convert moles of nitric acid to grams using its molar mass. The molar mass of HNO3 is approximately 63.01 g/mol, so:

Theoretical yield of HNO3 = moles of HNO3 × molar mass of HNO3

Theoretical yield of HNO3 = 2.0 moles × 63.01 g/mol = 126.02 grams

Thus, the theoretical yield of nitric acid when 2.0 moles of nitrogen dioxide react with enough oxygen is 126.02 grams.

Answer:

There are [tex]1.38506\times 10^{23} [/tex] of bromine molecules present in the flask.

Explanation:

Moles of liquid bromine in the flask = 0.230 mol

As we know that 1 mole is equal to the Avogadro number.

[tex]1 mol=6.022\times 10^{23}[/tex] atoms/ molecules

So, number of bromine molecules in 0.230 moles of liquid bromine is:

[tex]=0.230\times 6.022\times 10^{23}=1.38506\times 10^{23} molecules[/tex]

There are [tex]1.38506\times 10^{23} [/tex] of bromine molecules present in the flask.

The limiting reactant in a chemical reaction is determined by calculating the amount of product (AlBr3 in this case) that would be produced from the complete reaction of each of the reactants separately. The reactant yielding the lesser amount of product is the limiting reactant.

To determine the limiting reactant in a chemical reaction, you can use a method where you compare the amount of each reactant to the amount of product expected from its complete reaction. Taking the given equation, 2Al(s)+3Br2(g)→2AlBr3(s), we would calculate the amount of the product AlBr3 that would be formed from the complete reaction of each reactant, Al and Br2, separately. The reactant that yields the lesser amount of the product AlBr3 is the limiting reactant. This concept can be understood using a simple analogy. Let's think of making a sandwich: 2 slices of bread + 1 slice of cheese yields 1 sandwich. If you have an excess of bread but only one slice of cheese, you can only make one sandwich. The cheese is the limiting reactant as it limits the yield of sandwich and the amount of bread left unreacted is the excess reactant. Similarly, in the chemical reaction given above, the reactant that produces less amount of product (AlBr3) acts as the limiting reactant.

https://brainly.com/question/33417913

#SPJ11

Answer:

A. Carbon

Explanation:

Carbon is one of the most important elements for the structure of living things. It is responsible for 19% of the body composition of animals, behind only oxygen, which contributes 61%. The head is one of the main elements that make up the tissues and organs of animals.

Carbon has a special property: it can make four different chemical bonds. This means that, in addition to being able to bond with atoms of other elements such as hydrogen, oxygen, nitrogen, sulfur and phosphorus, carbon can also bond in different ways with other carbon atoms.

The result is the formation of thousands of carbon compounds that are present in nature, rocks, minerals, products made in industries, plastics, smoke, oil, plants, animals and within our own bodies.

Final answer:

The empirical formula of the compound is CH2O.

Explanation:

To determine the empirical formula of a substance, we need to find the ratio of the number of atoms of each element in the compound.

Step 1: Convert the given masses of each element to moles. Using the molar mass of each element (C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol), we have 5.28 g C = 0.44 mol C, 0.887 g H = 0.877 mol H, and 3.52 g O = 0.22 mol O.

Step 2: Divide each element's mole value by the smallest mole value to get the simplest ratio. In this case, the smallest mole value is 0.22 mol O. Dividing the other mole values by 0.22 gives us 0.44 mol C, 0.877 mol H, and 1 mol O.

The empirical formula of the compound is CH2O.

The atomic core for a potassium ion (K+) is [Ar]4s¹ without the one electron it loses to become an ion. The atomic core for a chloride ion (Cl-) is [Ne]3s² 3p⁵ with an extra electron gained to fill its 3p orbital.

The atomic cores for potassium and chloride ions are represented by their respective electron configurations and their charges after ion formation. Potassium (K) has an atomic number of 19, which means it has 19 electrons. The electron configuration is [Ar]4s¹. When potassium forms an ion, it loses one electron to achieve a noble gas configuration, making it a potassium ion with a charge of +1, represented as K⁺.

Chlorine (Cl) has an atomic number of 17, which means it has 17 electrons. The electron configuration is [Ne]3s² 3p⁵. When chlorine forms an ion, it gains one electron to fill its 3p orbital, becoming a chloride ion with a charge of -1, represented as Cl⁻.

https://brainly.com/question/9092255

#SPJ3

One Valence electron

Final answer:

To excel in radiology, Wade needs to develop strong interpersonal skills to communicate effectively with patients and excellent technical skills for using and understanding complex radiological equipment and procedures. So the correct options are C and D.

Explanation:

For Wade to excel as a radiologist, there are two essential skills that are particularly important. One of these skills is interpersonal skills, which would allow him to communicate effectively with patients, help them to relax, and explain procedures to provide a comfortable experience. The other key skill is technical skills relating to the use of radiological machinery and equipment. These skills are necessary for accurately performing and interpreting medical imaging, such as computed tomography (CT), magnetic resonance imaging (MRI), and mammography which involves complex operation of medical devices and analysis of the resulting images for diagnostic purposes.

While leadership skills and public speaking abilities can be beneficial in any profession, they are not as central as interpersonal and technical skills for the day-to-day responsibilities of a radiologist. Physical strength is generally less critical, given that lifting heavy equipment is not a common task for radiologists.