CHEMISTRY HIGH SCHOOL

when 1.00 mol of no( g ) forms from its elements, 90.29 kj of heat is absorbed. how kj much heat is evolved when 2.79 g of no decomposes to its elements? assistance

Answers

Answer 1
Answer: Molar Mass of the NO = 14 g/mol + 16 g/mol = 30 g/mol

Number of mols is 2.79 g = 2.79mol/30g/mol = 0.093 mol

Heat released by the decomposition of 0.093 mol = 92.29kj/mol * 0.093 mol = 8.58 kj (you might include a negative sign to indicate tha the heat is released)

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Can someone help me with my chemistry homework?!

Answers

Answer:

calculate grams and use mole ratio lol

Explanation:

Answer:

Explanation:

23

A 4.00 mL aliquot of a 0.15 M HCl solution is diluted to a final volume of 25.00 mL. What is the molarity of this first dilution solution? Then a second dilution was made by taking 7.50 mL of the first dilution and diluting it to 50.00 mL. What is the molarity of this second dilution?

Answers

Final answer:

The molarity of HCl after the first dilution is 0.024 M and after the second dilition, it is 0.0036M.

Explanation:

The problem at hand is a solution dilution problem in the field of Chemistry, usually tackled by the use of formula M1V1=M2V2, where M1 and V1 are the original molarity and volume, and M2 and V2 are the molarity and volume after dilution. In the first dilution, applying this formula gives (0.15 M)(4.00 mL) = (M2)(25.00 mL), solving for M2 gives a value of 0.024 M. The second dilution would similarly have the equation (0.024 M)(7.5 mL) = (M3)(50.0 mL), which gives M3 approximately 0.0036 M. Hence, the molarity of the first dilution would be 0.024 M and the molarity of the second dilution would be 0.0036 M.

Learn more about Solution Dilution here:

brainly.com/question/31521767

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I really dont know but it would be useful to use chegg it really help you find the answer in a proper manner

How many Significant Figures does 105.60 have? *

Answers

Answer:

5 significant figures

Explanation:

Which two substances bind using a lock-and-key mechanism?enzyme and substrate
reactant and substrate
substrate and product
reactant and product

Answers

The answer is the first option. Enzyme and substrate bind using a lock-and-key mechanism. Enzymes act on a specific substrate and a substrate needs a specific enzyme, this is what is called a lock-and-key mechanism. Enzymes and substtates are like a key and a lock, one is for each other.
it would be A: enzyme and substrate

The rate law for a hypothetical reaction is rate = k [A]2. When the concentration is 0.10 moles/liter, the rate is 2.7 × 10-5 M*s-1. What is the value of k?

Answers

Answer:The value of rate constant is 2.7* 10^(-3) Ls^(-1) mol^(-1).

Explanation:

Concentration of [A]=0.10 mol/L

The rate of the reaction = 2.7* 10^(-5) M/s

Rate constant = k

The given rate law:

R=k[A]^2

2.7* 10^(-5) M/s=k* (0.10 mol/L)^2

k=(2.7* 10^(-5) mol/L s)/(0.10 mol/L* 0.10 mol/L)=2.7* 10^(-3) Ls^(-1) mol^(-1)

The value of rate constant is 2.7* 10^(-3) Ls^(-1) mol^(-1).

Given:

rate = k [A]2

concentration is 0.10 moles/liter

rate is 2.7 × 10-5 M*s-1

Required:

Value of k

Solution:

rate = k [A]2

2.7 × 10-5 M*s-1 = k (0.10 moles/liter)^2

k = 2.7 x 10^-3 liter per mole per second
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