The balanced equation for the reaction between sulfur (S₈) and oxygen (O₂) to form sulfur trioxide (SO₃) is 2S₈ + 16O₂ → 16SO₃.
What is the balanced chemical equation?Balancing chemical equations involves the addition of stoichiometric coefficients to the reactants and products.
The balanced equation for the reaction between sulfur (S₈) and oxygen (O₂) to form sulfur trioxide (SO₃) is determined as;
2S₈ + 16O₂ → 16SO₃
From the reactants side we can see that sulfur is 16 and also 16 in the product side. The number of oxygen in the reactant side is 32 and also 32 in the product side.
Thus, the balanced equation for the reaction between sulfur (S₈) and oxygen (O₂) to form sulfur trioxide (SO₃) is 2S₈ + 16O₂ → 16SO₃.
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How many grams of NaOH are needed to make 100. mL of solution with a concentration of 1.5 M?
To create 100 mL of solution with a concentration of 1.5 M, 6.00 grams of NaOH are required.
The amount of NaOH needed to make 100. mL of solution with a concentration of 1.5 M can be calculated using the formula:
mass = molarity x volume x molar mass
where:
molarity = 1.5 M (given)
volume = 100. mL = 0.1 L (given)
molar mass of NaOH = 40.00 g/mol (from periodic table)
Substituting the values, we get:
mass = 1.5 mol/L x 0.1 L x 40.00 g/mol
mass = 6.00 g
Therefore, 6.00 grams of NaOH are needed to make 100. mL of solution with a concentration of 1.5 M.
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A rectangular ingot of gold is 23.7 cm long by 75.5 mm wide by 10.9 cm high. If 1.0 cm³ of gold weighs 19.30 g, what is the price of the ingot in SA Rand if the current price of gold is $1629.8 per ounce. (1oz = 28.35g Exchange rate = 9.6 R/$).
The price of the ingot in South African Rand (SAR) is 19,533,171.507 R.
To calculate the price of the ingot in South African Rand (SAR), we need to follow these steps:
Convert the dimensions of the ingot to cm³:
Length = 23.7 cm
Width = 75.5 mm = 7.55 cm
Height = 10.9 cm
The volume of the ingot is calculated by multiplying these dimensions:
Volume = Length × Width × Height
= 23.7 cm × 7.55 cm × 10.9 cm
= 1830.0475 cm³
Calculate the weight of the ingot in grams:
Since 1 cm³ of gold weighs 19.30 g, we can multiply the volume of the ingot by this conversion factor to obtain the weight:
Weight = Volume × 19.30 g
= 1830.0475 cm³ × 19.30 g
= 35,380.1375 g
Convert the weight of the ingot to ounces:
Since 1 ounce is equal to 28.35 g, we can divide the weight of the ingot by this conversion factor:
Weight in ounces = Weight / 28.35 g
= 35,380.1375 g / 28.35 g
= 1247.0461 ounces
Calculate the price of the ingot in USD:
The current price of gold is $1629.8 per ounce, so we can multiply the weight of the ingot in ounces by this price:
Price in USD = Weight in ounces × Price per ounce
= 1247.0461 ounces × $1629.8/ounce
= $2,032,881.72278
Convert the price from USD to SAR:
The exchange rate is 9.6 R/$, so we can multiply the price in USD by this exchange rate:
Price in ZAR = Price in USD × Exchange rate = $2,032,881.72278 × 9.6 R/$ = 19,533,171.507 R
Therefore, the price of the ingot in South African Rand (SAR) is approximately 19,533,171.507 R.
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WHAT IS THE PERCENT OF HYDROGEN IN CU(C2H3O2)2?
WITH SOLUTION
Answer:
Hydrogen H 3.330% Carbon C 26.450% Oxygen O 35.234%
Explanation:
Is it beneficial or harmful to man or both? Discuss how it is beneficial or harmful to man?
The crystal I chose is sodium chloride crystals and it is beneficial for man as it is used in the preservation of food as well as in seasoning of food.
What are crystals?A solid whose components are arranged in a highly ordered microscopic structure to form an all-pervasive crystal lattice is referred to as a crystal.
Sodium chloride also referred to as common salt is an ionic compound that has the chemical formula NaCl.
Sodium chloride is an essential nutrient employed in healthcare. It is employed as a spice to improve flavor and as a food preservative. Additionally, sodium chloride is employed in the production of plastics and other goods and is applied to de-ice sidewalks and roadways
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Which number is the same as 2.5
10-3?
The number that is the same as the exponentiation given as follows: 2.5 × 10-³ is 0.0025.
What is exponentiation?Exponentiation is the process of calculating a power by multiplying together a number of equal factors, where the exponent specifies the number of factors to multiply.
For example, if 10 is multiplied three times, then it can be written as "10 raised to 3" which means 10³. In this case, 10 is the base, and 3 is the exponent.
Therefore, a number 0.0025 can be written in exponentiation as 2.5 × 10-³ by counting the number of zeros forward.
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What is the energy associated with the formation of 2.55 g of 4He by the fusion of 3H and 1H?
Substance Mass (u)
4He 4.00260
3H 3.01605
1H 1.00783
The energy associated with the formation of 2.55 g of 4He by the fusion of 3H and 1H is -2.982 x 10⁻¹⁰ J.
The given masses of the isotopes can be converted to kilograms using the conversion factor: 1 u = 1.661 x 10⁻²⁷ kg.
Mass of 4He = 2.55 g = 2.55 x 10⁻³ kg
Mass of 3H = 3.01605 u = 3.01605 x 1.661 x 10⁻²⁷ kg/u
= 5.0099 x 10⁻²⁷ kg
Mass of 1H = 1.00783 u = 1.00783 x 1.661 x 10⁻²⁷ kg/u
= 1.6737 x 10⁻²⁷ kg
The balanced equation for the fusion reaction is;
3H + 1H → 4He
The molar mass of 4He is 4.0026 g/mol, which can be converted to kg/mol using the conversion factor: 1 g/mol = 1 x 10⁻³ kg/mol.
Molar mass of 4He = 4.0026 g/mol = 4.0026 x 10⁻³ kg/mol
The number of moles of 4He formed can be calculated from its mass;
n(4He) = m(4He) / M(4He)
= 2.55 x 10⁻³ kg / 4.0026 x 10⁻³ kg/mol
= 0.638 mol
From the balanced equation, 3 moles of H atoms react with 1 mole of He atoms to form 1 mole of He atoms. Therefore, the number of moles of H atoms required for the reaction is;
n(H) = 3/4 x n(4He)
= 3/4 x 0.638 mol
= 0.479 mol
The energy released in the reaction can be calculated using the mass-energy equivalence equation;
E = Δm c²
where Δm is change in mass, c is the speed of light.
The change in mass is;
Δm = [3H + 1H - 4He] = [5.0099 x 10⁻²⁷ kg + 1.6737 x 10⁻²⁷kg - 4.0026 x 10⁻³ kg]
= -3.315 x 10⁻²⁷ kg (negative because mass is lost in the reaction)
The energy released is;
E = (-3.315 x 10⁻²⁷ kg) c²
= (-3.315 x 10⁻²⁷ kg) (2.998 x 10⁸ m/s)²
= -2.982 x 10⁻¹⁰ J
The negative sign indicates that energy is released in the reaction (exothermic reaction).
Therefore, the energy associated is -2.982 x 10⁻¹⁰ J.
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If a solution has a [H+] concentration of 4.5 x 10-7 M, is this an acidic or basic solution?
Solve and Explain.
Considering the definition of pH, the pH is 6.35 and the solution is acidic.
Definition of pHpH is the Hydrogen Potential and it is a measure of acidity or alkalinity. pH indicates the amount of hydrogen ions present in a solution or substance.
Mathematically, pH is calculated as the negative base 10 logarithm of the activity of hydrogen ions:
pH= - log [H⁺]
The numerical scale that measures the pH of substances includes the numbers from 0 to 14. The pH value 7 corresponds to neutral substances. Acidic substances are those with a pH lower than 7, while basic substances have a pH higher than 7.
Acidic or basic solution in this caseIn this case, being [H⁺]=4.5×10⁻⁷ M, you can replace this value in the definition of pH:
pH= -log (4.5×10⁻⁷ M)
Solving:
pH= 6.35
Finally, the pH is lower than 7, the solution is acidic.
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2 NaN3 → 2 Na + 3 N
Given 9.98 grams of N2, how many moles of NaN3 are produced?
0.238 moles of NaN₃ are produced from 9.98 grams of N₂.
What is the moles of NaN₃ produced?The moles of he mass of NaN₃ produced
The balanced equation for the reaction is:
2 NaN₃ → 2 Na + 3 N₂
The molar ratio between NaN₃ and N₂ is 2:3, which means that for every 2 moles of NaN₃, 3 moles of N₂ are produced.
The mole ratio is used to determine how many moles of NaN₃ are produced from 9.98 grams of N₂.
First, we need to convert the mass of N₂ to moles:
moles of N₂ = mass of N2 / molar mass of N₂
moles of N₂ = 9.98 g / 28.02 g/mol
moles of N₂ = 0.356 mol
moles of NaN₃ = (2/3) * moles of N₂
moles of NaN₃ = (2/3) * 0.356 mol
moles of NaN₃ = 0.238 mol
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B. Answer the question.
How is living in space different from living on Earth?
Answer:
The three major differences are: atmosphere (vacuum in space), radiation (high level of dangerous particles), and gravity (weightlessness in space). The first difference between the Earth and space is the atmosphere.Jan
The only significant differences from living on Earth are that they operate in the confined space of the Space Shuttle orbiter cabin and that they, and all objects inside the cabin, float.
Explanation:
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A 25 L sample of oxygen gas (O2) has a mass of 48 grams and a pressure of 3.0 atm. What would be the temperature of the sample? Reminder: Use the equation PV=nRT, with the constant R = 0.0821 L atm/mol K.
A.
609 K
B.
305 K
C.
19.0 K
D.
1.60 x 10-2 K
The temperature of the oxygen gas sample is 609 K, which is approximately 336°C or 637°F. The answer is A.
We can use the ideal gas law equation, PV = nRT, to solve for the temperature of the oxygen gas sample.
First, we need to calculate the number of moles of oxygen gas present in the sample using its mass and molar mass:
n = m/M
where:
n = number of moles
m = mass (in grams)
M = molar mass (in g/mol)
The molar mass of oxygen gas (O2) is 32.00 g/mol.
n = 48 g / 32.00 g/mol = 1.50 mol
Next, we can rearrange the ideal gas law equation to solve for temperature (T):
T = (PV) / (nR)
where:
T = temperature (in Kelvin)
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L atm/mol K)
Plugging in the given values, we get:
T = (3.0 atm x 25 L) / (1.50 mol x 0.0821 L atm/mol K)
T = 609 K
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A sample of argon has a volume of 1.20 L at STP. If the temperature is increased to 28.0 °C and the pressure is lowered to 0.800 atm, what will the new volume be, in L?
The new volume of the sample of argon, in L, will be 1.43 L.
At STP (Standard Temperature and Pressure), the temperature is 0 °C (273 K) and the pressure is 1 atm. Therefore, the initial conditions can be expressed as:
T1 = 273 K
P1 = 1 atm
V1 = 1.20 L
To find the new volume, we can use the combined gas law:
(P1V1) / T1 = (P2V2) / T2
We can rearrange this equation to solve for V2:
V2 = (P1V1T2) / (P2T1)
Substituting the given values, we get:
V2 = (1 atm x 1.20 L x 301 K) / (0.800 atm x 273 K)
V2 = 1.43 L
At a temperature of 28.0 °C and a pressure of 0.800 atm, the new volume of the argon sample is 1.43 L.
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