What is the most abundant chemical element in the universe?

In which of the following does the land being moved contin approximately 60% water

sp2606 [1]

Answer:

C

Explanation:

The answer is C because only that amount can move

4 0

3 years ago

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Decreasing the temperature of the reaction 3H2 + N2 2NH3. In this reaction, the product absorbs heat. WHICH WAY WILL THE REACTIO

Alex

Answer:

the reaction will shift towards the “heat”—shifts to the left

Explanation:

To summarize:

o If temperature increases (adding heat), the reaction will shift away from the “heat” term and go in the

endothermic direction.

o If temperature decreases (removing heat), the reaction will shift towards the “heat” term and go in the

exothermic direction.

o NOTE: The endothermic direction is always away from the “heat” term and the exothermic direction is

towards the “heat” term.

Therefore the reaction will shift towards the “heat”—shifts to the left

3 0

3 years ago

The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c

netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is $4.58\times 10^{-3}M$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of oxygen gas

$C_2\text{ and }p_2$ are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used: 1 atm = 760 torr

$C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm$

Putting values in above equation, we get:

$\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M$

Hence, the solubility of oxygen gas at 628 torr is $4.58\times 10^{-3}M$

4 0

2 years ago

Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH

uranmaximum [27]

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

7 0

3 years ago

You gained two pounds while on vacation. Which of the following has changed?

Elenna [48]

I believe it is d. physical shape n particle amount (amount of particle increase if u gain weight right?), weight directly proportional to mass..

3 0

3 years ago

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