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3 years ago

If a chemical change has taken place, which of these is most likely true? It's for science.

Chemistry

2 answers:

Alexxandr [17]3 years ago

7 0

Answer:

A completely new substance if formed

Explanation:

Chemical changes occur when a substance combines with another to form a new substance, called chemical synthesis or, alternatively, chemical decomposition into two or more different substances. These processes are called chemical reactions and, in general, are not reversible except by further chemical reactions. Some reactions produce heat and are called exothermic reactions and others may require heat to enable the reaction to occur, which are called endothermic reactions. Understanding chemical changes is a major part of the science of chemistry.

leonid [27]3 years ago

4 0

Yes, it is most likely for science.

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Consider the balanced equation below. 4nh3 3o2 right arrow. 2n2 6h2o what is the mole ratio of nh3 to n2? 2:4 4:2 4:4 7:2

KIM [24]

The required mole ratio of NH₃ to N₂ in the given chemical reaction is 2:4.

<h3>What is the stoichiometry?</h3>

Stoichiometry of the reaction gives idea about the number of entities present on the reaction before and after the reaction.

Given chemical reaction is:

4NH₃ + 3O₂ → 2N₂ + 6H₂O

From the stoichiometry of the reaction it is clear that:

4 moles of NH₃ = produces 2 moles of N₂

Mole ratio NH₃ to N₂ is 2:4.

Hence required mole ratio is 2:4.

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3 years ago

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Calculate the number of moles in 8 g of water.

polet [3.4K]

Answer:

refer the above attachment

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2 years ago

How many grams of sodium chloride are contained in 250.0 g of a 15% NaCl solution?

kramer

Given concentration of NaCl=15%

Means ,

In every 100g of Solution 15g of NaCl is present .

Now

Given mass=250g

So ,

$\\ \Large\sf\longmapsto 250\times 15\%$

$\\ \Large\sf\longmapsto 250\times \dfrac{15}{100}$

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3 years ago

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Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

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3 years ago