Loops of gas that exit the sun is called what?

What is the the correct equation for water gas

nikdorinn [45]

Answer:

Water-gas A mixture of carbon monoxide (CO) and hydrogen (H2) produced by passing steam over red-hot coke using the endothermic reaction C + H2O → CO + H2.

6 0

3 years ago

When a 17.7 mL sample of a 0.368 M aqueous hypochlorous acid solution is titrated with a 0.301 M aqueous barium hydroxide soluti

nordsb [41]

Answer:

pH = 12.98

Explanation:

Step 1: Data given

Volume of aqueous hypochlorous acid solution = 17.7 mL = 0.0177 L

Molarity of aqueous hypochlorous acid solution = 0.368 M

Molarity of aqueous barium hydroxide solution = 0.301 M

Volume of aqueous barium hydroxide solution = 16.2 mL = 0.0162 L

Step 2: The balanced equation

2HCl + Ba(OH)2 → BaCl2 + 2H2O

Step 3: Calculate moles

Moles = molarity * volume

Moles HCl = 0368 M * 0.0177 L

Moles HCl = 0.0065136 moles

Moles Ba(OH)2 = 0.301 M * 0.0162 L

Moles Ba(OH)2 = 0.0048762 moles

Step 4: Calculate the limiting reactant

For 2 moles HCl we need 1 mol Ba(OH)2 to produce 1 mol BaCl2 and 2 moles H2O

HCl is the limiting reactant. It will completely be consumed 0.0065136 moles. Ba(OH)2 is in excess. There will react 0.0065136/2 = 0.0032568‬ moles. There will remain 0.0048762 moles - 0.0032568‬ = 0.0016194 moles

Step 5: Calculate molarity Ba(OH)2

Molarity Ba(OH)2 = moles / volume

Molarity Ba(OH)2 = 0.0016194 moles / 0.0339 L

Molarity Ba(OH)2 = 0.04777 M

Step 6: Calculate [OH-]

Ba(OH)2 → Ba^2+ + 2OH-

For Ba(OH)2 we have 2* [OH-]

[OH-] = 2*0.04777 = 0.09554 M

Step 7: Calculate pOH

pOH = -log[OH-]

pOH = -log(0.09554)

pOH = 1.02

Step 8: Calculate pH

pH = 14 - 1.02

pH = 12.98

8 0

3 years ago

Ethylene glycol (C2H6O2) is used as an antifreeze in cars. If 250 g of ethylene glycol is added to 3.00 kg of water, what is the

Zepler [3.9K]

Answer:

2,909 M

Explanation:

molair mass is of.ethylene is 26,04 g/mol

first you need to calculate how much mL 3 kg is. You can do this by using the density of ethylene: 1,1 g/mL.

3000 g x 1.1 = 3300 mL = 3,3 L

Next you need to calculate the amount of moles:

250 g / 26,04 g/mol = 9,60 mol

Now you can calculate the molarity:

9,6/3.3 = 2,909 M

I don't know the answer for the second question. I'm sorry.

3 0

3 years ago

Global Precipitation Measurement (GPM) is a tool scientists use to forecast weather. Which statements describe GPM? Select three

Simora [160]

Answer:

It is a satellite that collects data about rain and snow.

• Its orbit covers 90 percent of Earth's surface.

• The sensors measure microwaves.

Explanation:

Correct on edge

4 0

3 years ago

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7. The equilibrium constant Kc for the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 54.3 at 430°C. At the start of the reaction there are

Juli2301 [7.4K]

Answer:

[H2] = 0.0692 M

[I2] = 0.182 M

[HI] = 0.826 M

Explanation:

Step 1: Data given

Kc = 54.3 at 430 °C

Number of moles hydrogen = 0.714 moles

Number of moles iodine = 0.984 moles

Number of moles HI = 0.886 moles

Volume = 2.40 L

Step 2: The balanced equation

H2 + I2 → 2HI

Step 3: Calculate Q

If we know Q, we know in what direction the reaction will go

Q = [HI]² / [I2][H2]

Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Q =(n(HI)²) /(nH2 *nI2)

Q = 0.886²/(0.714*0.984)

Q =1.117

Q<Kc This means the reaction goes to the right (side of products)

Step 2: Calculate moles at equilibrium

For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI

Moles H2 = 0.714 - X

Moles I2 = 0.984 -X

Moles HI = 0.886 + 2X

Step 3: Define Kc

Kc = [HI]² / [I2][H2]

Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Kc =(n(HI)²) /(nH2 *nI2)

KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))

X = 0.548

Step 4: Calculate concentrations at the equilibrium

[H2] = (0.714-0.548) / 2.40 = 0.0692 M

[I2] = (0.984 - 0.548) / 2.40 = 0.182 M

[HI] = (0.886+2*0.548) /2.40 = 0.826 M

6 0

3 years ago

Read 2 more answers