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Murljashka [212]
3 years ago
8

What are tidal conditions in science terms

Chemistry
1 answer:
Anon25 [30]3 years ago
8 0
They are the tides and conditions of the oceans
You might be interested in
Any help would be appreciated. Confused.
masya89 [10]

Answer:

q(problem 1) = 25,050 joules;  q(problem 2) = 4.52 x 10⁶ joules

Explanation:

To understand these type problems one needs to go through a simple set of calculations relating to the 'HEATING CURVE OF WATER'. That is, consider the following problem ...

=> Calculate the total amount of heat needed to convert 10g ice at -10°C to steam at 110°C. Given are the following constants:

Heat of fusion (ΔHₓ) = 80 cal/gram

Heat of vaporization (ΔHv) = 540 cal/gram

specific heat of ice [c(i)] = 0.50 cal/gram·°C

specific heat of water [c(w)] = 1.00 cal/gram·°C

specific heat of steam [c(s)] = 0.48 cal/gram·°C

Now, the problem calculates the heat flow in each of five (5) phase transition regions based on the heating curve of water (see attached graph below this post) ...   Note two types of regions (1) regions of increasing slopes use q = mcΔT and (2) regions of zero slopes use q = m·ΔH.

q(warming ice) =  m·c(i)·ΔT = (10g)(0.50 cal/g°C)(10°C) = 50 cal

q(melting) = m·ΔHₓ = (10g)(80cal/g) 800 cal

q(warming water) = m·c(w)·ΔT = (10g)(1.00 cal/g°C)(100°C) = 1000 cal

q(evaporation of water) =  m·ΔHv = (10g)(540cal/g) = 5400 cal

q(heating steam) = m·c(s)·ΔT = (10g)(0.48 cal/g°C)(10°C) = 48 cal

Q(total) = ∑q = (50 + 800 + 1000 + 5400 + 48) = 7298 cals. => to convert to joules, multiply by 4.184 j/cal => q = 7298 cals x 4.184 j/cal = 30,534 joules = 30.5 Kj.

Now, for the problems in your post ... they represent fragments of the above problem. All you need to do is decide if the problem contains a temperature change (use q = m·c·ΔT) or does NOT contain a temperature change (use q = m·ΔH).    

Problem 1: Given Heat of Fusion of Water = 334 j/g, determine heat needed to melt 75g ice.

Since this is a phase transition (melting), NO temperature change occurs; use q = m·ΔHₓ = (75g)(334 j/g) = 25,050 joules.

Problem 2: Given Heat of Vaporization = 2260 j/g; determine the amount of heat needed to boil to vapor 2 Liters water ( = 2000 grams water ).

Since this is a phase transition (boiling = evaporation), NO temperature change occurs; use q = m·ΔHf = (2000g)(2260 j/g) = 4,520,000 joules = 4.52 x 10⁶ joules.

Problems containing a temperature change:

NOTE: A specific temperature change will be evident in the context of problems containing temperature change => use q = m·c·ΔT. Such is associated with the increasing slope regions of the heating curve.  Good luck on your efforts. Doc :-)

5 0
3 years ago
Molarity of 2%W/V of NaOH is
TiliK225 [7]
Assume it is 1 litre and weighs 1kg.
2 percent of 1 kg is 20g.
20g divided by molar mass of NaOH.
20g divide by 40 = 0.5 mole
0.5 mole in a litre would be 0.5M
That is the answer: 0.5M
6 0
3 years ago
each of the following reactions has been reported in the chemical literature. predict the product in each case, showing stereoch
poizon [28]

(a): This is an elimination reaction. The reaction proceeds in 3 steps; Step 1: Protonation of alcohol, step 2: Loss of leaving group, and step 3: Deprotonation to form an alkene. Elimination (E1) reaction dominates than substitution (SN1) reaction because the formed HSO4- anion is a poor nucleophile and tends not to add to the carbocation intermediate formed. Hence the product formed is I, is trans.

(b) This is a dihydroxylation of the alkenes reaction. A catalytic amount of OsO4 is used along with an oxidizing agent, which oxidizes the reduced osmium(VI) into osmium(VIII) to regenerate the catalyst. The reaction proceeds in 2 steps; Step 1: Cis addition of OsO4 to double bond and form 4 member cyclic osmate ester. Step 2: Hydrolysis of the intermediate to 1.2-diol. Hence the product formed is II, is cis. 

(c) This is hydroboration - oxidation reaction. The reaction proceeds through the addition of H- on the more substituted carbon and BH2 on the less substituted carbon of the double bond. The reaction is a syn addition. Hence the product formed is III.

(d) This a reduction reaction where carboxylic acid (-COOH) is reduced to 1o alcohol (-CH2OH). Hence the formed product is IV.

Stereochemistry, a subdiscipline of chemistry, involves the take a look at the relative spatial association of atoms that form the shape of molecules and their manipulation.

The observation of stereochemistry specializes in stereoisomers, which by means of definition have an equal molecular method and series of bonded atoms (constitution), however, differ within the three-dimensional orientations of their atoms in space. for that reason, it's also referred to as 3-d chemistry three-dimensionality.

Learn more about  Stereochemistry here:-brainly.com/question/13266152

#SPJ4

<u>Disclaimer:- your question is incomplete, please see below for the complete question.</u>

Each of the following reactions has been reported in the chemical literature. Predict the product in each case, showing stereochemistry where appropriate.

5 0
10 months ago
Give the symbols for metalloids
Elina [12.6K]

Answer:

The metalloids; boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), polonium (Po) and astatine (At) are the elements found along the step like line between metals and non-metals of the periodic table.

Elements: Germanium; Boron; Arsenic

Explanation:

6 0
3 years ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
2 years ago
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