Which of the following is similar among all living organisms on Earth?

ILL GIVE BRIANLY, FOLLOW , 5 STARS AND HEARTS

V125BC [204]

Answer:

it can help heat or cool a room.

Explanation:

I think this is right...

7 0

2 years ago

+ H₂O

trapecia [35]

Answer:

None of these are correct, because there is no way to balance this equation, but I hope these steps help you figure out your answer.

Explanation:

Count out the single amounts of elements you have on both sides of the equation. To be balanced, you need to have the exact same for each element.

Before balanced Left side.

Cl-2

O-8

H-2

Before balanced right side.

H-1

Cl-1

O-3

That means we need to increase Hydrogen, Chlorine and Oxygen on the right for sure and see how that affects the equation. You can keep adding the Coefficients until the # of elements begin to match on each side.

(I tried to balance this equation, it doesn't work, there is too much on the reactants side for what the product is.)

8 0

2 years ago

What happens when carbon dioxide is passed through lime water for a long time? write with equation

Vlada [557]

Answer:

when CO2 gas is passed through lime water it turns milky due to the formation of calcium carbonate which formula is CaCO3.

Ca(OH)2+ CO2------ CaCO3

when excess of carbon dioxide is passed through calcium carbonate calcium hydrogen carbonate is formed and solution become colourless.

CaCO3+CO2------ Ca(HCO3)

5 0

3 years ago

The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker. 1.

Mars2501 [29]

C: 0.012 mol.

<h3>Explanation</h3>

Start with 0.0020 moles of iodate ions ${\text{IO}_{3}}^{-}$ .

How many moles of iodine $\text{I}_2$ will be produced?

${\text{IO}_{3}}^{-}$ converts to $\text{I}_2$ in the first reaction. The coefficient in front of $\text{I}_2$ is three times the coefficient in front of ${\text{IO}_{3}}^{-}$ . In other words, each mole of ${\text{IO}_{3}}^{-}$ will produce three moles of $\text{I}_2$ . 0.0020 moles of ${\text{IO}_{3}}^{-}$ will convert to 0.0060 moles of $\text{I}_2$ .

How many moles of thiosulfate ions ${\text{S}_2\text{O}_3}^{2-}$ are required?

$\text{I}_2$ reacts with ${\text{S}_2\text{O}_3}^{2-}$ in the second reaction. The coefficient in front of $\text{I}_2$ is twice the coefficient in front of ${\text{S}_2\text{O}_3}^{2-}$ . How many moles of ${\text{S}_2\text{O}_3}^{2-}$ does each mole of $\text{I}_2$ consume? Two. 0.0060 moles of $\text{I}_2$ will be produced. As a result, $2 \times 0.0060 = 0.0120$ moles of ${\text{S}_2\text{O}_3}^{2-}$ will be needed.

6 0

2 years ago

Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163

statuscvo [17]

Answer:

ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

------------------------------------------------------------------------------------

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

5 0

3 years ago