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4 years ago

Is a type of physical weathering

2 answers:

4 0

There are two main types of physical weathering: Freeze-thaw occurs when water continually seeps into cracks, freezes and expands, eventually breaking the rock apart. Exfoliation occurs as cracks develop parallel to the land surface a consequence of the reduction in pressure during uplift and erosion.

ELEN [110]4 years ago

3 0

Answer:

abrasion

Explanation:

answer on apx

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Classify these salts as acidic, basic, or neutral. KCL, NH4Br, K2CO3, NaCN, LiClO4? This is what I put but got it wrong: KCl = b

MaRussiya [10]

In classifying salts as neutral, acidic, or basic, it is important to take note of the strength of the acids and bases that they come from. A strong acid and strong base produce a neutral salt. A weak acid and strong base produce a basic salt. A strong acid and weak base produce an acidic salt. So the answers must be:

KCl = neutral (from HCl and KOH)
NH4Br = acidic (from NH4 and HBr)
K2CO3 = basic (from KOH and H2CO3)
NaCN = basic (from NaOH and HCN)
LiClO = basic (from LiOH and HClO)

3 0

4 years ago

Read 2 more answers

A correct name for the following compound is:_________.a) 4-bromo-3,8-dimethylbicyclo[5.2.2]nonane b) 3,8-dimethyl-4-bromo-bicyc

mixer [17]

Answer:

e) 4-bromo-3,8-dimethylbicyclo[5.2.0]nonane

Explanation:

The missing image of the the compound we are to name is attached below.

Before we can name an organic compound; It is crucial we know the guiding rules in naming them.

1. Select the longest continuous carbon chain as the root hydrocarbon and name according to the number of carbon atoms it contains, adding appropriate suffix to indicate the principal substituent group.

2. Number the carbon atoms in the root hydrocarbon from the end which gives the lowest number to the substituents.

3. If the same substituent is present two or more times in a molecule; the number of this substituent is indicated by the prefix di -(2), tri - (3) , tetra - (4) etc attached to the substituent name.

4. If there is more than one type of substituent in the molecule ; the substituents are named according to the alphabetical order but where there are mixed substituents ; the inorganic are named first.

5. In selecting and numbering the longest continuous chain, the functional groups are given preference over substituents., i.e the functional group is given the smallest possible number.

In the light of the above guiding rules; we were able to name the given compound because the compound contains nine carbons in the ring form which result to root name nonane. The two methyl are on the third and eight carbon; bromine is on the fourth carbon ; there are two cyclic ring present in the compound where we have 5 carbons in one structure, another 2 carbons in the second structure and zero carbon in the bridge structure which eventually result to the correct name:

4-bromo-3,8-dimethylbicyclo[5.2.0]nonane

5 0

3 years ago

How many moles of ions are released when 0.27 mole of cobalt chloride is dissolved in water?

egoroff_w [7]

0.81 moles

Explanation:

cobalt chloride when dissolved in water dissociate into cobalt and chloride ions as shown;

CoCl₂ (s) → [Co²⁺](aq) + 2[Cl⁻] (aq)

The mole ratio between CoCl₂ and [Co²⁺] is 1 ; 1 while the mole ratio between CoCl₂ and [Cl⁻] is 1 : 2

Therefore; 0.27 moles of CoCl₂ will form;

(1 * 0.27) = 0.27 moles of [Co²⁺] ions

(2 * 0.27) = 0.54 moles of [Cl⁻] ions

Total moles of dissociated ions are;

0.27 + 0.54

= 0.81 moles

4 0

3 years ago

When 2.714 g of AX (s) dissolves in 127.4 g of water in a coffee-cup calorimeter the temperature rises from 23.3 °C to 37.1 °C.

MakcuM [25]

Answer : The enthalpy change for the solution is 166.34 kJ/mol

Explanation :

First we have to calculate the enthalpy change of the reaction.

Formula used :

$\Delta H=mC\Delta T\\\\\Delta H=mC(T_2-T_1)$

where,

$\Delta H$ = change in enthalpy = ?

C = heat capacity of water = $4.18J/g.K$

m = total mass of sample = 2.174 + 127.4 = 129.6 g

$T_1$ = initial temperature = $23^oC=273+23=296K$

$T_2$ = final temperature = $37.1^oC=273+37.1=310.1K$

Now put all the given values in the above expression, we get:

$\Delta H=mC(T_2-T_1)$

$\Delta H=129.6g\times 4.18J/g.K\times (310.1-296)K=7638.36J$

Now we have to calculate the moles of AX added to water.

$\text{ Moles of }AX=\frac{\text{ Mass of }AX}{\text{ Molar mass of }AX}=\frac{2.714g}{59.1097g/mole}=0.04592moles$

Now we have to calculate the enthalpy change for the solution.

As, 0.04592 moles releases heat = 7638.36 J

So, 1 moles releases heat = $\frac{7638.36}{0.04592}=166340.59J=166.34kJ$

Therefore, the enthalpy change for the solution is 166.34 kJ/mol

7 0

3 years ago

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T

motikmotik

Answer :

(a) The rate law for the reaction is:

$\text{Rate}=k[OCl^-]^1[I^-]^1$

(b) The value of rate constant is, $60.4M^{-1}s^{-1}$

(c) rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$OCl^-+I^-\rightarrow OI^-+Cl^-$

Rate law expression for the reaction:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

where,

a = order with respect to $OCl^-$

b = order with respect to $I^-$

Expression for rate law for first observation:

$1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(1)

Expression for rate law for second observation:

$2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(2)

Expression for rate law for third observation:

$2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1$

Dividing 1 from 3, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1$

Thus, the rate law becomes:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

a = 1 and b = 1

$\text{Rate}=k[OCl^-]^1[I^-]^1$

Now, calculating the value of 'k' (rate constant) by using any expression.

$1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})$

$k=60.4M^{-1}s^{-1}$

Now we have to calculate the rate for a reaction when concentration of $OCl^-$ and $I^-$ is $1.8\times 10^{-3}M$ and $6.0\times 10^{-4}M$ respectively.

$\text{Rate}=k[OCl^-][I^-]$

$\text{Rate}=(60.4M^{-1}s^{-1})\times (1.8\times 10^{-3}M)(6.0\times 10^{-4}M)$

$\text{Rate}=6.52\times 10^{-5}Ms^{-1}$

Therefore, the rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

8 0

3 years ago