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3 years ago

GUYS I NEED HELP WITH THIS ASSIGNMENT MY LAST ASSIGNMENT FOR THE DAY

Chemistry

1 answer:

Aleksandr-060686 [28]3 years ago

7 0

Answer:

the reactants are na and cl2 the products is the combination nacl --the law of conservation of matter keeps reactions balanced. No you would not be adding more atoms that is what the first combo is. The coefficient 2 stands for 2 atoms of KL

Explanation:

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Describe the method of steel manufacture used when removes the impurities carbon, phosphorus and sulfur

melisa1 [442]

Explanation:

Steelmaking is the process of producing steel from iron ore and/or scrap. In steelmaking, impurities such as nitrogen, silicon, phosphorus, sulfur and excess carbon (most important impurity) are removed from the sourced iron, and alloying elements such as manganese, nickel, chromium, carbon and vanadium are added to produce different grades of steel.

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6 0

3 years ago

A chemist prepares a solution of copper(II) sulfate CuSO4 by measuring out 31.μmol of copper(II) sulfate into a 150.mL volumetri

baherus [9]

Answer:

The concentration of the copper (II) sulfate solution is 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM

Explanation:

The concentration of a solution is the amount of solute dissolved in a given volume of solution. In this case, the concentration of the copper(II) sulfate solution in micromoles per liter (symbol ) is the number of micromoles of copper(II) sulfate dissolved in each liter of solution. To calculate the micromoles of copper(II) sulfate dissolved in each liter of solution you must divide the total micromoles of solute by the number of liters of solution.

Here's that idea written as a formula: c= n/V

where c stands for concentration, n stands for the total micromoles of copper (II) sulfate and V stands for the total volume of the solution.

You're not given the volume of the solution in liters, but rather in milliliters. You can convert milliliters to liters with a unit ratio: V= 150. mL * 10^-3 L/ 1 mL = 0.150 L

Next, plug in μmol and liters into the formula to divide the total micromoles of solute by the number of liters of solution: c= 31 μmol/0.150 L = 206.66 μmol/L

Convert this number into scientific notation: 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM

3 0

2 years ago

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

2 years ago

What does it mean for a strong base to be in equilibrium?

prisoha [69]

The position of equilibrium lies far to the right, with products being favoured. Hence, option A is correct.

<h3>What is equilibrium?</h3>

Chemical equilibrium is a condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs.

A very high value of K indicates that at equilibrium most of the reactants are converted into products.

The equilibrium constant K is the ratio of the concentrations of products to the concentrations of reactants raised to appropriate stoichiometric coefficients.

When the value of the equilibrium constant is very high, the concentration of products is much higher than the concentration of reactants.

This means that most of the reactants are converted into products and the position of equilibrium lies far to the right, with products being favoured.

Hence, option A is correct.

Learn more about the equilibrium here:

brainly.com/question/23641529

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8 0

2 years ago

Rank the following atoms by number of valence electrons. Rank from most to fewest valence electrons. To rank items as equivalent

Fed [463]

To count the number of valence electrons we look at the electronic configuration and add the electrons form the electronic shell with the highest principal quantum number.

Rb: [Kr] 5s¹ - 1 valence electron

Xe: [Kr] 5s² 4d¹⁰ 5p⁶ - 8 valence electrons

Sb: [Kr] 5s² 4d¹⁰ 5p³ - 5 valence electrons

I: [Kr] 5s² 4d¹⁰ 5p⁵ - 7 valence electrons

In: [Kr] 5s² 4d¹⁰ 5p¹ - 3 valence electrons

Rank from most to fewest valence electrons:

Xe > I > Sb > In > Rb

6 0

3 years ago

Read 2 more answers