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muminat
3 years ago
7

What is ductility in a solid?

Chemistry
1 answer:
Elodia [21]3 years ago
4 0

Answer:

D

Explanation:

REASONS WHY THE REST ARE WRONG

1. For A, that's only in when non-conductivity of solid is required.

2. For B, that's when malleability of the solid is required.

3. For C, that's when the conductivity of solid is required

4. D is talking about the ductility of solids

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Igneous rocks would most likely be found near:
geniusboy [140]

Answer:

the deep sea floor. Known as the oceanic crust.

Explanation:

The deep seafloor (the oceanic crust) is made almost entirely of basaltic rocks, with peridotite underneath in the mantle. Basalts are also erupted above the Earth's great subduction zones, either in volcanic island arcs or along the edges of continents.

Hope this helps :)

7 0
3 years ago
What parts are present in every chemical equation?
WITCHER [35]
The reactants, products, coefficients, subscripts. ( I forgot the rest lol)
5 0
2 years ago
Give examples which indicate that nylon fibres are very strong.​
Readme [11.4K]

Explanation:

is used for making ropes, used for climbing rocks and for making parachutes. Their usage shows that nylon fibres have high tensile strength

3 0
3 years ago
Determine the change in volume that takes place when a 2.00 L sample of N2 gas is heated from 250 C to 500 C.
Effectus [21]

Answer:

V₂ = 2.96 L

Explanation:

Given data:

Initial volume = 2.00 L

Initial temperature =  250°C

Final volume = ?

Final temperature = 500°C

Solution:

First of all we will convert the temperature into kelvin.

250+273 = 523 k

500+273= 773 k

According to Charles's law,

V∝ T

V = KT

V₁/T₁ = V₂/T₂

V₂ = T₂V₁/T₁

V₂ = 2 L × 773 K / 523 k

V₂ =  1546 L.K / 523 k

V₂ = 2.96 L

4 0
3 years ago
The gas phase decomposition of sulfuryl chloride at 600 K SO2Cl2(g)SO2(g) + Cl2(g) is first order in SO2Cl2. During one experime
krok68 [10]

Answer: The rate constant for the reaction is 3.96\times 10^{-3}min^{-1}

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample = 559 min

a = let initial amount of the reactant = 2.83\times 10^{-3}

a - x = amount left after decay process  = 3.06\times 10^{-4}

559min=\frac{2.303}{k}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}

k=\frac{2.303}{559}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}

k=3.96\times 10^{-3}min^{-1}

The rate constant for the reaction is 3.96\times 10^{-3}min^{-1}

6 0
3 years ago
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