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3 years ago

Identify the variables in this hypothesis.

Chemistry

2 answers:

musickatia [10]3 years ago

7 0

Answer:

pressure and volume

Explanation:

Zanzabum3 years ago

5 0

Answer:

AJJJJJJJJJJJJJJ

Explanation:

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CFC-11 is typically the CFC with the shortest atmospheric lifetime.

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5 What is a physical property of matter?

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Explanation:

Characteristic of matter that is not associated with its change in chemical composition.

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3 years ago

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HELP ME PLEASE (brainliest)

Aloiza [94]

Answer: i would say D or the last one.

Explanation: According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants. The law of conservation of mass is useful for a number of calculations and can be used to solve for unknown masses, such the amount of gas consumed or produced during a reaction.

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3 years ago

What is 10 times as many as 1 hundred is blank hundred or blank thousand ?

Rama09 [41]

10 times as many as 1 hundred is 10 hundred or 1 thousand.

This is the question of simple multiplications related to a place value system.

The given equation in unit form can be rewritten as:

10 × 1 × 100 = (X × 100) or (Y × 1000)

Arranging in two parts, we get,

10 × 1 × 100 = X × 100 ........(1)

10 × 1 × 100 = Y × 1000 ........(2)

Now, calculating the value of X and Y,

From equation (1), we get,

10 × 1 × 100 = X×100

or, 1000 = X × 100

or, X = 10

Hence, 10 times as many as 1 hundred is 10 hundred.

Similarly, from equation (2). we get,

10 × 1 × 100 = Y × 1000

or, 1000 = Y × 1000

or, Y = 1

Hence, 10 times as many as 1 hundred is 1 thousand.

Therefore, 10 times as many as 1 hundred is 10 hundred or 1 thousand.

To learn more about hundreds, thousands, and place values, visit: brainly.com/question/17785584

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8 0

1 year ago

Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---> CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu

Nitella [24]

ΔH = +438 kJ

We have three equations:

(I) N₂ + 3H₂ → 2NH₃; ΔH = -92 kJ

(II) H₂ +½O₂ → H₂O; ΔH = -286 kJ

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; ΔH = -632 kJ

From these, we must devise the target equation:

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = ?

_________________________________

The target equation has 2NH₃ on the left, so you reverse equation (I).

When you reverse an equation, you reverse the sign of its ΔH.

(V) 2NH₃ → N₂ + 3H₂; ΔH = +92 kJ

Equation (V) has 1N₂ on the right, and that is not in the target equation.

You need an equation with 1N₂ on the left.

Reverse Equation (III).

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

Equation (VI) has ³/₂O₂ on the right, and that is not in the target equation.

You need ³/₂O₂ on the left.

Multiply Equation (II) by three.

When you multiply an equation by three, you multiply its ΔH by three.

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

_______________________________________

We get the target equation (IV):

(V) 2NH₃ → N₂ + 3H₂; ΔH = + 92 kJ

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = +438 kJ

7 0

3 years ago