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2 years ago

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Rosa was looking for patterns to help predict the products of chemical reactions. She recorded three similar decomposition react

ions in the table.
A 2-column table with 3 rows. The first column labeled reactants has entries 2 N a C l O subscript 3, 2 K C l O subscript 3, 2 L i C l O subscript 3. The second column labeled products has entries 2 N a C l + 3 O subscript 2, 3 O subscript 2 + 2 K C l, empty.

What products should she record in the last row of the table?

2 answers:

kipiarov [429]2 years ago

5 0

The products should she record in the last row of the table : <u>2LiCl + 3O₂</u>

<h3>Further explanation</h3>

There are several chemical reactions, namely:

Formation reaction
Decomposition reaction.
Replacement reaction.
Multiple replacement reactions
Neutralization reaction.
Combustion reaction.
Polymerization

The decomposition reaction in a chemical reaction shows the decomposition reaction of a compound into its constituent elements or compounds

Rosa recorded three similar decomposition reactions

Reactions that occur :

1. 2 NaClO₃ ⇒ 2NaCl + 3O₂

2. 2KClO₃ ⇒ 2KCl + 3O₂

3. 2LiClO₃ ⇒ 2LiCl + 3O₂

There are similarities in the decomposition that forms oxygen and chloride compounds

Learn more

homogenous mixture of two or more pure substances

brainly.com/question/1832352

Keywords : products, patterns, chemical reaction, table, decomposition reactions

#LearnWithBrainly

boyakko [2]2 years ago

4 0

Answer:

2LiCl + 3O₂

Explanation:

Hopefully this helps

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At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci

kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

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A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga

andreev551 [17]

Answer:

Given that

V= 0.06 m³

Cv= 2.5 R= 5/2 R

T₁=500 K

P₁=1 bar

Heat addition = 15000 J

We know that heat addition at constant volume process ( rigid vessel ) given as

Q = n Cv ΔT

We know that

P V = n R T

n=PV/RT

n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

So

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

$\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}$

Cp-Cv= R

Cp=7/2 R

Now by putting the values

$\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}$

$\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}$

a)ΔS= 20.79 J/K

b)

If the process is adiabatic it means that heat transfer is zero.

So

ΔS= 20.79 J/K

We know that

$\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}$

Process is adiabatic

$\Delta S_{surr}=0$

$\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}$

$\Delta S_{univ}= 20.79 +0$

$\Delta S_{univ}= 20.79$

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you want to compare brands of paper towels to see which holds the most liquid. the independent variable in your experiment would

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Answer: the brand of paper towel

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Kiting during a storm. The legend that Benjamin Franklin flew a kite as a storm approached is only a legend — he was neither stu

Dvinal [7]

Answer:

The current is $I = 1.1434*10^{-5}}\ A$

Explanation:

From the question we are told that

The radius of the kite string is $R = 2.02 mm = 0.00202 \ m$

The distance it extended upward is $D = 0.823 km = 823 \ m$

The thickness of the water layer is $d = 0.506 mm = 0.000506 \ m$

The resistivity is $\rho = 159\ \Omega \cdot m$

The potential difference is $V = 186 MV = 186 *10^{6} \ V$

Generally the cross sectional area of the water layer is mathematically represented as

$A = \pi r^2$

Here r is mathematically represented as

$r = [(R + d ) - R]$

=> $r = [(0.00202 + 0.000506 ) - 0.00202]$

=> $r = 0.000506$

=> $A = 3.142 * [0.000506]^2$

=> $A = 8.0447*10^{-7}\ m^2$

Generally the resistance of the water is mathematically represented as

$R = \frac{\rho * D }{A}$

=> $R = \frac{159 *823 }{8.0447*10^{-7}}$

=> $R = 1.62662 * 10^{11} \ \Omega$

Generally the current is mathematically represented as

$I = \frac{V}{R}$

=> $I = \frac{186 *10^{6} }{1.62662 * 10^{11}}$

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2 years ago