10. Chromium (III) oxide reacts with carbon dioxide to form..

3. 79,789kJ = (293k) (377kJ)

KatRina [158]

the anwser would most likely be j=0

5 0

3 years ago

Which structures join with the cell’s membrane during exocytosis?

erma4kov [3.2K]

Answer:Exocytosis is also used to integrate new proteins into the cell membrane. In this process, the new protein is formed inside the cell, and migrates to phospholipid bilayer of the vesicle. The vesicle, containing the new protein as a part of the phospholipid bilayer, fuses with the cell membrane.

Explanation:

8 0

2 years ago

In fall, leaves may change from green to yellow or red. Explain in your own words what is happening inside the leaf with regard

natima [27]

Answer:

The pigment that causes leaves to be green is chlorophyll. ... As chlorophyll goes away, other pigments start to show their colors. This is why leaves turn yellow or red in fall. In fall, plants break down and reabsorb chlorophyll, letting the colors of other pigments show through.

3 0

2 years ago

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

3 years ago

How do I figure out how to solve the table ?

nataly862011 [7]

You have to add a photo to we can understand - Yuno Gasai

3 0

3 years ago

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