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Soloha48 [4]
2 years ago
15

Balance these equations! 1) AIBr; + K- KBr + AI

Chemistry
1 answer:
Tema [17]2 years ago
3 0

Answer:

Instructions on balancing chemical equations:

Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below

Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F.     Compare: Co - cobalt and CO - carbon monoxide

To enter an electron into a chemical equation use {-} or e

To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}.

Example: Fe{3+} + I{-} = Fe{2+} + I2

Substitute immutable groups in chemical compounds to avoid ambiguity.

For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,

but PhC2H5 + O2 = PhOH + CO2 + H2O will

Compound states [like (s) (aq) or (g)] are not required.

If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.

Reaction stoichiometry could be computed for a balanced equation. Enter either the number of moles or weight for one of the compounds to compute the rest.

Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reage

Explanation:

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Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT
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(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

(c) L(T, α) = 60·α·T - 9000·α + 60

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L =  L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

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L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

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