I really need help with this before 5:10

Water difference between glacier and geyser

Gekata [30.6K]

Glaciers are found in areas where the net accumulation of snow exceeds the melt, therefore they are usually found in areas of high elevation, cold temperatures, and high precipitation.

Geysers are found in areas of shallow crust with high heat flow, such as hot spots.

4 0

3 years ago

Read 2 more answers

The compound sodium thiosulfate pentahydrate, Na2S2O3 • 5H2O

Mnenie [13.5K]

The theoretical yield is 204.4 g while the percent yield is 2.57%.

<h3>What is theoretical yield?</h3>

Theoretical yield is the amount of product obtained based on the stoichiometry of the reaction.

S8(s) + 8 Na2SO3(aq) + 40 H2O(l) --->8 Na2S2O3·5 H2O(s)

Number of moles of sulfur = 3.25 g /8(32) = 0.013 moles

Number of moles of sodium sulfite = 13.1 g/126 g/mol = 0.103 moles

Since 1 moles of sulfur reacts with 8 moles of sodium sulfite

0.013 moles reacts with 0.013 moles × 8 moles /1 mole = 0.104 moles

There is not enough sodium sulfite hence it is the limiting reactant.

1 mole of sodium sulfite yields 8 moles of product

0.103 moles of sodium sulfite yields 0.103 moles × 8 moles /1 mole = 0.824 moles

Mass of product = 0.824 moles × 248 g/mol = 204.4 g

percent yield = 5.26 g /204.4 g × 100/1

= 2.57%

Learn more about percent yield: brainly.com/question/2506978

4 0

2 years ago

A flask contains a mixture of hydrogen gas and water vapor at STP. If the pressure of the water vapor is 19.5 mmHg, then what is

horrorfan [7]

Answer:

I would say A but not sure

Explanation:

7 0

3 years ago

Why is the answer C for this problem?

pshichka [43]

Answer:

$\boxed{\text{(C) X}$_{3}$P$_{2}}$

Explanation:

Step 1. Identify the Group that contains X

We look at the consecutive ionization energies and hunt for a big jump between them

$\begin{array}{crc}n & IE_{n} & IE_{n} - IE_{n-1}\\1 & 730 & \\2 & 1450 & 720\\3 & 7700 & 6250\\4 & 10500 & 2800\\\end{array}$

We see a big jump between n = 2 and n = 3. This indicates that X has two valence electrons.

We can easily remove two electrons, but the third electron requires much more energy. That electron must be in the stable, filled, inner core.

So, X is in Group 2 and P is in Group 15.

Step 2. Identify the Compound

X can lose two valence electrons to reach a stable octet, and P can do the same by gaining three electrons.

We must have 3 X atoms for every 2 P atoms.

The formula of the compound is $\boxed{\text{X}$_{3}$P$_{2}}$}$ .

4 0

3 years ago

You are asked to prepare 500 mL 0.300 M500 mL 0.300 M acetate buffer at pH 4.904.90 using only pure acetic acid ( MW=60.05 g/mol

defon

Answer:

You will need 9,0 g of acetic acid

Explanation:

The equilibrium acetate-acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76

Using Henderson-Hasselbalch you will obtain:

pH = pka + log₁₀ $\frac{[A^{-}]}{[HA]}$

Where HA is acetic acid and A⁻ is acetate ion

4,90 = 4,76 + log₁₀ $\frac{[A^{-}]}{[HA]}$

1,38 = $\frac{[A^{-}]}{[HA]}$ <em>(1)</em>

As acetate concentration is 0,300M:

0,300M = [HA] + [A⁻] <em>(2)</em>

Replacing (2) in (1):

[HA] = 0,126 M

And:

[A⁻] = 0,174 M

As you need to produce 500 mL:

0,5 L × 0,126 M = 0,063 moles of acetic acid

0,5 L × 0,174 M = 0,087 moles of acetate

To produce moles of acetate from acetic acid:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:

0,087 moles of acetate + 0,063 moles of acetic acid ≡ 0,15 moles of acetic acid × $\frac{60,05 g}{1mol}$ = <em>9,0 g of acetic acid</em>

I hope it helps!

5 0

3 years ago