if we react 10 moles of hydrogen with 6 moles of oxygen. Oxygen is the limiting reagent and 10moles of water is produced
The limiting reagent is the reactant that runs out first, limiting the amount of product that can be produced. In order to determine the limiting reagent, we need to compare the stoichiometry of the reactants with the balanced equation. We can see that for every 2 moles of H2, 2 moles of H2O are produced. Also for every 1 mole of O2, 2 moles of H2O are produced.To determine the limiting reagent we have to find out which reactant is in excess, for that we can divide the number of moles of each reactant by its coefficient in the balanced equation. 10 moles of H2 / 2 moles of H2/1 mole of H2 = 5 moles of H2O are produced 6 moles of O2 / 2 moles of H2O/1 mole of O2 = 3 moles of H2O are producedSince 3 moles of H2O are produced, we can see that oxygen (O2) is the limiting reagent. So we can't produce more than 3 moles of H2O and the amount of water produced is 3 moles.
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write a nuclear equation for the fusion of h−3 with h−1 to form he−4.
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This reaction releases a large amount of energy, which is what makes it useful in nuclear fusion reactors.
What is nuclear fusion?Nuclear fusion is a process in which two or more atomic nuclei join together to form a single, heavier nucleus.
The nuclear equation for the fusion of h-3 with h-1 to form he-4 is:
3H-1 + 1H-3 -> 4He-2 + 1n0
This equation describes the fusion of two hydrogen atoms to form a helium atom and a neutron. The first hydrogen atom (H-1) has a mass of 1, while the second hydrogen atom (H-3) has a mass of 3. When the two hydrogen atoms fuse together, they form a helium atom (He-2) with a mass of 4 and a neutron (n0) with a mass of 1. This reaction releases a large amount of energy, which is what makes it useful in nuclear fusion reactors.
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does any solid cu(oh)2 form when 0.075 g of koh is dissolved in 1.0 l of 1.0 × 10−3 m cu(no3)2?
Yes, the precipitation of Cu(OH)₂ will form when 0.075 g of KOH is dissolved in 1.0 L of 1.0 X 10⁻³ M Cu(NO₃)₂.
Chemical reactions begin with reactants as well as products of a reaction are the substances that were produced. The common chemical formula can be used to represent a chemical reaction: Reaction products. During the chemical reactions, bonds shatter and reform. Also, the opposite outcome of the reaction will happen.
[OH⁻]₀ = 0.075g/56.105g/mol(1.01) = 1.34×10⁻³
Cu(OH)₂ (s) ⇄ Cu²⁺ (aq) + 2OH⁻(aq)
? 0.0010 1.34×10⁻³
Q = [Cu²⁺][OH⁻]² = 0.0010 ×(1.34×10⁻³)² = 1.8×10⁻⁹ > Ksp 2.2×10⁻²⁰
This means that equilibrium will shift to the left, i.e., to reactants in which in turn implies that solid Cu(OH)₂ will be formed.
moles of Cu²⁺= 1.L x 1 x 10⁻³ M= 10⁻³
moles of KOH = 0.075 g/ 56.107 g/mol=0.00133
[Cu²⁺]= 10⁻³ / 1. L= 10⁻³ M
[OH⁻]=0.00133 / 1.L= 0.00133 M
Qsp= 10⁻³ (0.00133)² =1.786 x 10⁻⁹ >> Ksp ( = 2.2 x 10⁻²⁰)
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0.25 moles of a gas at 760 mmHg
and 298 K are contained in a 6.1 L
bottle. What is the pressure of the
system if the amount of gas in the
bottle is reduced to 0.13 mole and
the temperature is reduced to 100 K?
Answer: 2.3*10^3 Pa.
Explanation: We can use the ideal gas law to determine the pressure of the system: PV = nRT
where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
The ideal gas constant R = 8.314 J/mol*K
Given:
n1 = 0.25 moles
V1 = 6.1 L
T1 = 298 K
P1 = 760 mmHg
n2 = 0.13 moles
V2 = 6.1 L
T2 = 100 K
We can use the equation PV = nRT to find the new pressure P2, using the information provided:
P2 = (n2RT2)/V2
We know that the volume of the bottle remains the same and we can convert the pressure from mmHg to Pascals
P2 = (0.138.314100)/6.1 = 2.3*10^3 Pa
Therefore, the pressure of the system when the amount of gas in the bottle is reduced to 0.13 mole and the temperature is reduced to 100 K is 2.3*10^3 Pa.
Answer:
132.62
Explanation:
i got the answer correct in the program
what is the name and formula of the chemical reagent used to separate the cations of group 1 from ions in other groups.
When water K2CrO4 is added, a yellow precipitate of PbCrO4 forms, confirming the presence of Pb2+ in the aqueous solution.
What does grouping cations into compounds mean?1. Grouping the cations into various categories. The anions supplied by the group reagents are used to precipitate the cations of each succeeding group as compounds. The precipitate from one group's cations is separated (usually by centrifugation followed by decantation).
In Group 1 Group 2 and transition metals, how are cations formed?More electropositive elements, such as those from groups 1, 2, 13, d-block (transition metals), and actinides, typically form cations. The oxidation states of D-block elements are positive, and they can form cations.
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The concept that electrostatic repulsion between electron pairs surrounding an atom causes these pairs to beseparated as far as possible is the foundation ofa. VSEPR theory. b. the hybridization model. c. the electron-sea model.d. Lewis theory.
The concept that the electrostatic repulsion between electron pairs surrounding an atom causes these pairs to be separated as far as possible is the foundation of VSEPR theory.
The premise of VSEPR is that valence pairs surrounding atoms tend to repel each other, so we adopt configurations that minimize this repulsion. This reduces the molecule's energy, increases its stability, and determines the shape of the molecule. The main assumptions of VSEPR theory are:
The shape of a molecule depends on the number of electron pairs in the valence shell around the central atom. ii The electron pairs in the valence shell repel each other because the electron cloud is negatively charged. VSEPR theory is used to predict the shape of a molecule from the electron pairs surrounding the central atom of the molecule.
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a chemist has 70 ml of 50% methane solution. how much of an 80% siolution must she add yo get a 60% solution?
a chemist has 70ml of 50% methane solution.8.75 ml of 80% methane solution to the 70 ml of 50% methane solution to get a 60% solution.
To solve this problem, you can use the formula for mixing two solutions of different concentrations:
C1V1 + C2V2 = C3V3
Where:
C1 = concentration of the first solution
V1 = volume of the first solution
C2 = concentration of the second solution
V2 = volume of the second solution
C3 = final concentration of the mixture
V3 = final volume of the mixture
We know that:
C1 = 50% = 0.50 (as a decimal)
V1 = 70 ml
C3 = 60% = 0.60 (as a decimal)
So we can rearrange the formula to solve for V2:
V2 = (C3V3 - C1V1) / C2
Since we don't know the final volume V3, we can assume it to be V1+V2
V2 = (0.60 * (V1 + V2) - 0.50 * V1) / 0.80
Solving this equation we have,
V2 = (0.60 * (70 + V2) - 0.50 * 70) / 0.80
V2 = (42 - 35) / 0.80
V2 = 7/0.8
V2 = 8.75 ml
So the chemist needs to add 8.75 ml of 80% methane solution to the 70 ml of 50% methane solution to get a 60% solution.
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determine the mass of ammonium nitrate (in g) that has the same number of nitrogen atoms as 2.5 liters of liquid nitrogen (n2). density of liquid nitrogen is 0.808 g/ml.
The mass of ammonium nitrate (in g) that has the same number of nitrogen atoms as 2.5 liters of liquid nitrogen (n2) is 72.10g.
Given the volume of liquid nitrogen (V) = 2.5L = 2500ml
The liquefied form of the element nitrogen, known as liquid nitrogen, is created commercially through fractional distillation of liquid air.
The density of liquid nitrogen is (d) = 0.808 g/ml.
Let the mass of ammonium nitrate = m
We know that density = mass/volume
mass = density x volume
The molecular weight of liquid nitrogen = 28.0134g
m = (0.808 x 2500) x 1 mole N2/28.0134 = 72.10g
Hence the required mass of ammonium nitrate is 72.10g
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calculate the concentration (% m/m) of nacl solution that was made by dissolving 40.0 g of sodium chloride in enough water to make 500.0 g of solution g.
A solution with an 8% (m/m) concentration was created by dissolving the sodium chloride in 40.0 g in 500.0 g of water. Sodium chloride, also known as table salt.
Sea salt has the chemical formula NaCl, which indicates a 1:1 ratio of sodium chloride ions, and is an ionic material despite also containing other chemical salts. With molar weights of 22.99 and 35.45 g/mol, respectively, 39.34 g of Na and 60.66 g of Cl make up 100 g of NaCl. The concentration can apply to any type of chemical mixture, even though it is most frequently used to describe solutes and solvents in solutions. Two variations of the molar (amount) concentration are the normal concentration and the osmotic concentration.
(40.0/500)*100, which equals 8, is the concentration formula.
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