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irga5000 [103]
3 years ago
8

After a hypothesis is tested a scientist must do what next ?

Chemistry
1 answer:
zubka84 [21]3 years ago
3 0
Test the hypothesis in an expirement!
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Complete the table for ion charge based upon their losing or gaining electrons in the outer shell. (Use the periodic table as ne
PolarNik [594]

Answer:

Explanation:

Group one:

The elements of group one shows +1 charge because these all are metals and lose their one valance electron.

Hydrogen lithium sodium potassium rubidium cesium francium

Group 2:

The elements of group two shows +2 charge because these all alkali metals and lose their two valance electrons.

beryllium magnesium calcium strontium barium radium

Group 3:

The elements of group three-B shoes +3 charge by losing three valance electrons.

Scandium yttrium lanthanum actinium

Group 4:

The elements of group 4th A and 4th B lose four electrons or gain four electrons to complete the octet and shows +4 or -4 charge.

Group 5:

Group 5th elements gain three electrons and shows -3 charge to complete the 8 electrons. (octet).

It involve the elements of group 5th A.

Group 6:

The elements of group 6A gain two electrons to complete the octet and shows -2 charge.

Group 7:

The elements of group 7A gain one electron to complete the octet and shows -1 charge.

Group 8:

The elements of group 8A are noble gases and have complete octet. That's why shows 0 charge.

4 0
2 years ago
Which of the following statements is true?
vaieri [72.5K]
I think D?? I apologize if not-check in other answers to be sure ^^
7 0
3 years ago
Read 2 more answers
Name two metalloids that are semiconductors
guajiro [1.7K]

The Metalloids are:

Boron.

Silicon.

5 0
3 years ago
Similarities between pluton and pegmatite
snow_lady [41]
Plutons are large chambers of magma under grown

pegmatites generally form in pluton so it cools slow enough to make the crystals big enough to be classified as pegmatite and not just granite
7 0
3 years ago
Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT
marysya [2.9K]

Answer:

(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

(c) L = L₀ + ΔL  = L₀ +  α × L₀ × ΔT = L₀ +  α × L₀ × (T - T₀)

Therefore;

L = 60 +  α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

7 0
3 years ago
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