Every organism is an ecosystem receives energy from another source.Which of these organism receives its energy directly from the sun? A-Grass C-deer B-fungus D-hawk

Answer:

55J

Explanation:

∆E = 400-345 = 55J

Answer:

Energy dissipated as sound is 55 J

Explanation:

Here we have the principle of conservation of energy which states that energy can neither be created nor destroyed but can be transformed from one form to another

Where there is an initial 400 J of thermal energy in the heating element of the electric kettle we have;

Total available energy = 400 J

The energy (heat) transferred to the water is given as 345 J

The heat dissipated as sound of the kettle during heating is then found as follows;

Total available energy = Heat transferred to water + Energy dissipated as sound

400 = 345 + Energy dissipated as sound

∴ Energy dissipated as sound = 400 - 345 = 55 J.

Answer:

P₂ = 1.0 atm

Explanation:

Boyles Law problem => P ∝ 1/V at constant temperature (T).

Empirical equation

P ∝ 1/V => P = k(1/V) => k = P·V => for comparing two different case conditions, k₁ = k₂ => P₁V₁ = P₂V₂

Given

P₁ = 1.6 atm

V₁ = 312 ml

P₂ = ?

V₂ = 500 ml

P₁V₁ = P₂V₂ => P₂ = P₁V₁/V₂ =1.6 atm x 312 ml / 500ml = 1.0 atm

Answer:

Base

Explanation:

acid + base = salt + water

The compound reacts with an acid to form a salt and water is base.

A base is a substances that is slippery to touch,corrosive and sour taste which react with acid to give salt and water. It turns red lithmus paper to blue. It is the degree of hydroxide ion in a solution.

The compound reacts with an acid to form a salt and water is base.

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Answer : The final volume of gas will be, 26.3 mL

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 0.974 atm

[tex]P_2[/tex] = final pressure of gas = 0.993 atm

[tex]V_1[/tex] = initial volume of gas = 27.5 mL

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]22.0^oC=273+22.0=295K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]15.0^oC=273+15.0=288K[/tex]

Now put all the given values in the above equation, we get:

[tex]\frac{0.974 atm\times 27.5 mL}{295K}=\frac{0.993 atm\times V_2}{288K}[/tex]

[tex]V_2=26.3mL[/tex]

Therefore, the final volume of gas will be, 26.3 mL

Answer:

volume of gas will be, 26.3 mL

Explanation:

Answer:

Limiting reactant is H2.

Excess reactant is O2.

Maximum 5.6 mol of H2O can be produced.

Explanation:

                               2H2 + O2 ----> 2 H2O

from reaction         2 mol   1 mol    

given                    5.6mol   4.7mol

calculated            5.6mol   2.8 mol

We can see that  for 5.6 mol H2 only 2.8 mol O2 needed, but we have 4.7 mol O2 given, so we have excess of O2.

Then limiting reactant is H2.

Excess reactant is O2.

                               2H2 + O2 ----> 2 H2O

from reaction         2 mol                2 mol

given                      5.6 mol             x mol = 5.6 mol

Final answer:

The maximum number of moles of H₂O that can be produced from 5.6 mol H₂ and 4.7 mol O₂ is 5.6 mol H₂O, with hydrogen (H₂) as the limiting reactant and oxygen (O₂) as the excess reactant.

Explanation:

To answer the question of the maximum number of moles of H₂O that can be produced from the reaction of 5.6 mol H₂ and 4.7 mol O₂, we must first look at the balanced chemical equation for the reaction which is 2H₂ + O₂ ightarrow 2H₂O. From this equation, we can see that every 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

Now, let's see if we have enough of each reactant:

Hydrogen: 5.6 moles H₂ is available

Oxygen: 4.7 moles O₂ is available

According to the stoichiometry of the equation, oxygen will run out first since 4.7 moles of oxygen can react completely with (4.7 times 2) = 9.4 moles of hydrogen. But only 5.6 moles of hydrogen are available, which is less than 9.4 moles, so actually, hydrogen will limit the reaction.

Thus, the limiting reactant is hydrogen (H₂) and the excess reactant is oxygen (O₂). The maximum number o2f moles of water that can be produced is therefore equal to the moles of hydrogen available, which is 5.6 moles of H₂, resulting in 5.6 moles of H₂O being produced.

Answer:

the rectives are just not doing anything so just do sodium

Explanation:

Answer:

D. Atmospheres

Explanation:

Answer:

4L

Explanation:

No of mole= concentrate × volume

Volume = n/C

V= 10/2.5 = 4L

Answer: 0

Explanation:

The oxidation number for nitrogen in NH3 (ammonia) is -3, determined by setting up the equation x + 3(+1) = 0, where x represents the oxidation number of nitrogen.

The oxidation number for NH3 (ammonia) can be found by considering the usual oxidation states of nitrogen and hydrogen. Hydrogen typically has an oxidation number of +1, except when it forms hydride compounds with metals. Since ammonia consists of one nitrogen atom and three hydrogen atoms, and since each hydrogen has an oxidation number of +1, the total oxidation number contributed by the hydrogen atoms is +3 (3 x +1).

To find the oxidation number of nitrogen in NH3, let the oxidation number be represented as x. The sum of the oxidation numbers in a neutral compound is zero. Therefore, if we have x as the oxidation number of nitrogen and +3 from the hydrogen atoms, we can set up the equation x + 3(+1) = 0 to solve for x. Simplifying, we get x = -3. Thus, the oxidation number of nitrogen in NH3 is -3.

Following this approach, we can understand how oxidation numbers reflect the degree of electron transfer between atoms in a chemical compound or during a chemical reaction, such as NH3 reacting with O2 to form N2 and H2O. The reaction 4 NH3 + 3 O2 → 2 N2 + 6 H2O illustrates this electron transfer process and the involvement of oxidation states in balancing chemical equations.

Answer:

3.1 *10^22 atoms Cu

Explanation:

M(Cu) = 63.5 g/mol

3.3 g Cu * 1mol Cu/63.5 g Cu = 0.0520 mol Cu

0.0520 mol Cu * 6.02*10^23 atoms Cu/1 mol Cu = 3.1 *10^22 atoms Cu

Answer:

false, false, true

Answer:

James Watson

Explanation:

ΔH° = -769.2 kJ is required.

Explanation:

Given equations are:

Pb(s) + PbO₂(s) + 2 H₂SO₄(l) → 2 PbSO₄(s) + 2H₂O(l);  ΔH° = –509.2 kJ ------1

SO₃(g) + H₂O(l) → H₂SO₄(l);  ΔH° = –130. kJ ----2

From the above 2 equations, by adding or subtracting or multiplying or dividing the required amount to get the final equation by means of Hess's law.

Multiplying eq. 2 by 2 we will get,

2 SO₃(g) + 2 H₂O(l) → 2 H₂SO₄(l) ;  ΔH° = -260 kJ ----3

Adding it to eq. 1 we will get,

Pb(s) + PbO₂(s) + 2 H₂SO₄(l) → 2 PbSO₄(s) + 2H₂O(l);  ΔH° = –509.2 kJ ------1

2 H₂SO₄ and 2 H₂O gets cancelled since they are on opposite sides, and the ΔH° values are added to get the ΔH° value of the required equation as,

Pb(s) + PbO₂(s) + 2 SO₃(g) → 2 PbSO₄(s)

ΔH° = -509.2 - 260 = -769.2 kJ

Answer:

The answer to your question is 1795.5 grams of CO₂

Explanation:

Data

mass of carbon dioxide = ?

mass of Octane = 571.3 g

Process

1.- Write the balanced chemical reaction

         C₈H₁₆ + 12O₂   ⇒  8CO₂  +  8H₂O

2.- Calculate the molar mass of octane and carbon dioxide

Octane = (12 x 8) + (16 x 1) =96 + 16 = 112 g

Carbon dioxide = 8[ 12 + 2(16)] = 352 g

3.- Use proportions to find the mass of carbon dioxide

              112 g of Octane ---------------- 352 g of carbon dioxide

              571.3 g of Octane -------------  x

                    x = (571.3 x 352) / 112

                    x = 201097.6 / 112

                   x = 1795.5 grams of CO₂

Answer:

The energy released as heat when 9.94 g Cu 2 O ( s ) undergo oxidation at constant pressure is -10.142 kJ

Explanation:

Here we have

2Cu₂O ( s ) + O₂ ( g ) ⟶ 4 CuO ( s ) Δ H ∘ rxn = − 292.0 kJ mol

In the above reaction, 2 Moles of Cu₂O (copper (I) oxide) react with one mole of O₂ to produce 4 moles of CuO, with the release of − 292.0 kJ/mol of energy

Therefore,

1 Moles of Cu₂O (copper (I) oxide) react with 0.5 mole of O₂ to produce 2 moles of CuO, with the release of − 146.0 kJ  of energy

We have 9.94 g of Cu₂O with molar mass given as 143.09 g/mol

Hence the number of moles in 9.94 g of Cu₂O is given as

9.94/143.09 = 6.95 × 10⁻² moles of Cu₂O

6.95 × 10⁻² moles of Cu₂O will therefore produce 6.95 × 10⁻² ×  − 146.0 kJ mol  or -10.142 kJ.

Final answer:

The energy released as heat when 9.94 g of copper(I) oxide undergoes oxidation is calculated by determining the moles of Cu2O from its mass and using the stoichiometry of the reaction along with its enthalpy change to find the energy change. The result is approximately -10.14 kJ.

Explanation:

The problem involves calculating the energy released as heat during the oxidation of copper(I) oxide (Cu2O) to copper(II) oxide (CuO), given the mass of copper(I) oxide and the enthalpy change of the reaction. The reaction is 2 Cu2O(s) + O2(g) → 4 CuO(s) with an enthalpy change (ΔH°rxn) of -292.0 kJ/mol.

To find the energy released, we first need to find the number of moles of Cu2O. The molar mass of Cu2O is 143.09 g/mol (2*63.55 + 15.99). Using the given mass of Cu2O (9.94 g), we have:

moles of Cu2O = mass (g) / molar mass (g/mol) = 9.94 g / 143.09 g/mol = 0.0694 mol

The reaction stoichiometry shows that 2 moles of Cu2O yield -292.0 kJ, so we need to calculate the energy for 0.0694 moles:

Energy released = 0.0694 mol * (-292.0 kJ/mol) / 2 = -10.14 kJ

Therefore, the energy released as heat when 9.94 g of Cu2O undergoes oxidation at constant pressure is approximately -10.14 kJ.

Volumetric flask are calibrated to contain.

A volumetric flask is calibrated to contain a specific volume of solution when filled to its calibration mark, while pipettes are designed to deliver precise volumes of liquids.

A volumetric flask is calibrated 'to contain' (T. C.) a specific volume of solution rather than to dispense it. This means when the flask is filled to its calibration mark, it is accurate to a specific volume such as 10.00 mL ± 0.02 mL for a 10-mL volumetric flask or 250.0 mL ± 0.12 mL for a 250-mL volumetric flask. Unlike volumetric flasks, pipettes like volumetric and graduated pipettes are designed to deliver a known volume of liquid, either a single volume in the case of volumetric pipettes or variable volumes for graduated pipettes. To achieve accurate measurements, it is essential that both pipets and volumetric flasks are clean because any residue can affect the volume of liquids either delivered or contained.

Answer:

11.4 moles of H₂SO₄ are needed to completely react the 7.6 moles of Al

Explanation:

The equation indicates that 2 moles of aluminum react to 3 moles of sulfuric acid in order to produce 1 mol of aluminum sulfate and 3 moles of hydrogen gas.

The reaction is:  2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

This question can be solved with an easy rule of three. Ratio in the reaciton is 2:3, so we propose:

2 moles of Al react with 3 moles of sulfuric acid

Then, 7.6 moles of Al will react with 11.4 moles of H₂SO₄

Answer:

N₂ = 0.7515atm

O₂ = 0.1715atm

NO = 0.0770atm

Explanation:

For the reaction:

N₂(g) + O₂(g) ⇄ 2NO(g)

Where Kp is defined as:

[tex]Kp = \frac{P_{NO}^2}{P_{N_2}P_{O_2}}}[/tex]

Pressures in equilibrium are:

N₂ = 0.790atm - X

O₂ = 0.210atm - X

NO = 2X

Replacing in Kp:

0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]

0.0460 = 4X² / 0.1659 - X + X²

0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²

-3.954X² - 0.0460X + 7.6314x10⁻³ = 0

Solving for X:

X = - 0.050 → False answer. There is no negative concentrations.

X = 0.0385 atm → Right answer.

Replacing for pressures in equilibrium:

N₂ = 0.790atm - X = 0.7515atm

O₂ = 0.210atm - X = 0.1715atm

NO = 2X = 0.0770atm

Answer:

partial pressure N2 =  0.7515 atm

partial pressure O2 =  0.1715 atm

partial pressure NO =  0.077 atm

Explanation:

Step 1: Data given

Temperature = 2200 °C

Pressure of N2 = 0.790 atm

Pressure of O2 = 0.210 atm

Kp = 0.0460

Step 2: The balanced equation

N2(g) + O2(g) ⇆ 2NO(g)

Step 3: The pressure at equilibrium

pN2 = 0.790 - X atm

pO2 = 0.210 - X atm

pNO = 2X

Step 4: Define Kp and the partial pressures

Kp = (pNO)² / (pO2 * pN2)

0.0460 = 4X² / (0.210 - X)(0.790 - X)

X = 0.0385

pN2 = 0.790 - 0.0385 =  0.7515 atm

pO2 = 0.210 - 0.0385 = 0.1715 atm

pNO = 2*0.0385 = 0.077 atm

Answer:

[tex]\large \boxed{\text{121 L}}[/tex]

Explanation:

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.700 atm

n = 3.80 mol

T = -1  °C

Calculations:

1. Convert the temperature to kelvins

T = (-1 + 273.15) K= 272.15 K

2. Calculate the volume

[tex]\begin{array}{rcl}pV &=& nRT\\\text{0.700 atm} \times V & = & \text{3.80 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{272.15 K}\\0.700V & = & \text{84.86 L}\\V & = & \textbf{121 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{121 L}}$}[/tex]

Answer: 24.6kpa

Explanation:

This is a classic case where Boyle's Law can be applied.

The equation for Boyle's Law is given as: P1V1 = P2V2

Where P1 = initial pressure

P2 = final pressure

V1 = initial volume

V2 = final volume

From the question, P1 = ?, P2 = 98.2kpa , V1 =4L , V2 =1L

P1V1 = P2V2

P1 x 4L = 98.2 kpa x 1L

Make P1 subject of formula we then have:

P1 = 98.2kpa X 1L / 4L

= 24.6kpa

Answer:

24.6kpa

Explanation:

Answer:C

Explanation:

Answer:

Radioactive

Explanation:

Just did this on study island

The AH for the reaction is 1043 kJ/mol and the reaction is endothermic.

The reaction you provided:

2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

AH = -948 kJ/mol - (-1311 kJ/mol + -286 kJ/mol + -394 kJ/mol)

AH = -948 kJ/mol + 1311 kJ/mol + 286 kJ/mol + 394 kJ/mol

AH = 1043 kJ/mol

The value of AH is 1043 kJ/mol, which means the reaction is endothermic. This is because the AH value is positive, indicating that energy is absorbed from the surroundings during the reaction.

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To find the ΔH for the decomposition of sodium bicarbonate, we subtract the sum of the enthalpies of formation of the reactants from those of the products, which results in -95 kJ. This indicates the reaction is exothermic.

To determine the enthalpy change (ΔH) for the decomposition of sodium bicarbonate (baking soda), we use the given enthalpies of formation for the reactants and products in the reaction:

2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(l)

The enthalpy (ΔH) for the reaction is calculated as follows:

ΔH = { ΔHf(products) - ΔHf(reactants) }

For this reaction, the enthalpy change is:

ΔH = [(-1311 kJ/mol for Na2CO3) + (-394 kJ/mol for CO2) + (-286 kJ/mol for H2O)] - [2 x (-948 kJ/mol for NaHCO3)]

ΔH = (-1311 - 394 - 286) - (2 x -948)

ΔH = -1991 + 1896

ΔH = -95 kJ

Since the enthalpy change is negative, we can classify this reaction as exothermic. An exothermic reaction is one that releases heat to the surroundings, as evidenced by the negative sign of ΔH.

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Answer:

9.03*10^23 atoms  of C

Explanation:

1 mol of any substance contains 6.02*10^23 particle of this substance.

1 mol C ---    6.02*10^23 atoms of C

1.50 mol C ---  x atoms of C

x = 1.50*6.02*10^23 = 9.03*10^23 atoms of C

Answer:

9L

Explanation:

Given parameters:

Initial volume V₁ = 3.6L

Initial pressure P₁  = 2.5atm

Final pressure P₂  = 1atm

Unknown:

Final volume V₂  = ?

Condition: constant temperature  = 25°C

Solution:

This problem compares the volume and pressure of a gas at constant temperature.

This is highly synonymous to the postulate of Boyle's law. It states that "the volume of a fixed mass of gas is inversely proportional to the pressure provided that temperature is constant".

Mathematically;

              P₁V₁  = P₂V₂

where P and V are pressure and volume

          1 and 2 are initial and final states

Input the parameters and solve for V₂;

          2.5 x 3.6  = 1 x V₂

                  V₂  = 9L

Final answer:

When the volume of a chamber containing CH₄ increased from 7.00L to 8.20L, 0.260 moles of CH₄ were added, assuming constant temperature and pressure.

Explanation:

To determine how many moles of CH₄ (methane) were added when the volume of the chamber increased from 7.00L to 8.20L, we can use the molar volume concept under the assumption that temperature and pressure remain constant, therefore following Avogadro's Law. According to Avogadro's Law, equal volumes of gases at the same temperature and pressure contain an equal number of moles. You can calculate the number of moles in the new volume by setting up a proportion based on the known conditions (3.00 L contains 0.650 moles) and then solving for the number of moles in the new volume of 8.20 L.

First, determine the number of moles in the 7.00 L chamber:

moles in 3.00 L / 3.00 L = moles in 7.00 L / 7.00 L

0.650 moles / 3.00 L = x moles / 7.00 L

x = (0.650 moles / 3.00 L) × 7.00 L

x = 1.517 moles in 7.00 L

Now, calculate the number of moles in the increased volume of 8.20 L:

moles in 3.00 L / 3.00 L = moles in 8.20 L / 8.20 L

0.650 moles / 3.00 L = y moles / 8.20 L

y = (0.650 moles / 3.00 L) × 8.20 L

y = 1.777 moles in 8.20 L

The number of moles added is the difference between the moles in 8.20 L and the moles in 7.00 L:

moles added = 1.777 moles - 1.517 moles

moles added = 0.260 moles

Therefore, when the volume of the CH₄ chamber increased from 7.00L to 8.20L, 0.260 moles of CH₄ were added.

Option B is correct. The statement that describes all solids is that they have a definite shape and volume

States of matter are one of the ways in which matter exists. Matter can exist as a solid, liquid, gas, and plasma.

The states of matter have different characteristics. Some of the properties of solids are:

The liquid contains loosely packed atoms eliminating the first option. Based on the explanations above, we can conclude that solids have a definite shape and volume.

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I believe that leading zeros, and tail zeros without a decimal point are not significant

Answer: 5.039moles

Explanation:

No. of moles=mass/molar mass

= 362.7/72= 5.039moles

Molar mass of C5H12= 72

Final answer:

To find the number of moles of C5H12 in 362.8 grams, divide the mass by the molar mass of C5H12 (72.0 g/mol), resulting in 5.04 moles.

Explanation:

To calculate how many moles of C5H12 are in 362.8 grams of the compound, you first need to determine the molar mass of C5H12. The molar mass of a compound is calculated by summing the molar masses of all atoms in the molecule. For C5H12, the calculation is as follows:

Carbon (C) has a molar mass of 12.0 g/mol, and there are 5 carbon atoms, which contributes 5 * 12.0 = 60.0 g/mol.Hydrogen (H) has a molar mass of 1.0 g/mol, and there are 12 hydrogen atoms, which contributes 12 * 1.0 = 12.0 g/mol.

Therefore, the molar mass of C5H12 is 60.0 g/mol (carbon) + 12.0 g/mol (hydrogen) = 72.0 g/mol.

To find the number of moles, divide the mass of the compound by its molar mass:

Moles of C5H12 = Mass of C5H12 / Molar mass of C5H12 = 362.8 g / 72.0 g/mol = 5.04 moles.

So, there are 5.04 moles of C5H12 in 362.8 grams of the compound.

Answer:

The concentration of the solution is 6 [tex]10^{-4}[/tex] mol/dm3.

Explanation:

We have the relationship between concentration and volume to be:

[tex]c_{1} v_{1} = c_{2} v_{2}[/tex]   where c1v1 and c2v2 are initial and final concentrations and volumes respectively.

NOTE: (volume should be in litres if concentration is in Molar, M. But seeing as you have 2 × 10-3, I assume you have converted from mL to L already.)

15mL × 0.002mol/dm3 = 50mL × [tex]x[/tex] mol/dm3

∴ [tex]x[/tex] = [tex]\frac{0.03}{50}[/tex]

= 0.0006 ≅ 6 ×[tex]10^{-4}[/tex]mol/dm3

Based on the data provided, the final concentration of the solution is  6 * 10⁻⁴ mol/dm3

Concentration of a solution is the amount of solute dissolved in a given volume of solution in litres.

The formula: C1V1 = C2V2 is used to calculate the concentration of solutions after dilution where

Using C1V1 = C2V2:

15 × 0.002= 50 × C2

C2 = 6 * 10⁻⁴ mol/dm3

Therefore, the final concentration of the solution is  6 * 10⁻⁴ mol/dm3.

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Answer:

The answer to your question is    V = 0.32 L

Explanation:

Data

Volume of NH₃ = ?

P = 3.2 atm

T = 23°C

mass of CaH₂ = 2.65 g

Balanced chemical reaction

               6Ca  +  2NH₃   ⇒   3CaH₂  +  Ca₃N₂

Process

1.- Convert the mass of CaH₂ to moles

-Calculate the molar mass of CaH₂

 CaH₂ = 40 + 2 = 42 g

                             42 g ------------------ 1 mol

                              2.65 g --------------  x

                              x = (2.65 x 1)/42

                              x = 0.063 moles

2.- Calculate the moles of NH₃

                     2 moles of NH₃ --------------- 3 moles of CaH₂

                      x                        --------------- 0.063 moles

                                x = (0.063 x 2) / 3

                                x = 0.042 moles of NH₃

3.- Convert the °C to °K

Temperature = 23°C + 273

                      = 296°K

4.- Calculate the volume of NH₃

-Use the ideal gas law

              PV = nRT

-Solve for V

                V = nRT / P

-Substitution

                V = (0.042)(0.082)(296) / 3.2

-Simplification

               V = 1.019 / 3.2

-Result

               V = 0.32 L