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jok3333 [9.3K]
2 years ago
15

Formation of water: 2H2 + 1 O2 --> 2H2O

Chemistry
1 answer:
noname [10]2 years ago
7 0

Answer:

2.2 moles H2O

Explanation:

35g O_2 \mbox{ \cdot }\frac{1mol}{32g/mol} \mbox{ \cdot }\frac{2mol H_2O}{1mol O_2}= 2.1875, which rounds to about 2.2

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The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab. How many moles of dextrose is this equivalent t
Katena32 [7]

The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab the moles of dextrose is this equivalent to is 3.6888 moles.

<h3>What are moles?</h3>

A mole is described as 6.02214076 × 1023 of a few chemical unit, be it atoms, molecules, ions, or others. The mole is a handy unit to apply due to the tremendous variety of atoms, molecules, or others in any substance.

To calculate molar equivalents for every reagent, divide the moles of that reagent through the moles of the restricting reagent. The calculation is follows:

  • 655/12 x 6 + 12+ 16 x 6
  • = 655/ 180 = 3.6888 moles.

Read more about moles:

brainly.com/question/24322641

#SPJ1

6 0
2 years ago
What is the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure? As air
Marina86 [1]

We need to know the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure.

The relationship is: As air pressure in an area increases, the density of the gas particles in that area increases.

For any gaseous substance, density of gas is directly proportional to pressure of gas.

This can be explained from idial gas edquation:

PV=nRT

PV=\frac{w}{M}RT [where, w= mass of substance, M=molar mass of substance]

PM=\frac{w}{V}RT

PM=dRT [where, d=density of thesubstance]

So, for a particular gaseous substance (whose molar mass is known), at particular temperature, pressure is directly related to density of gaseous substance.

Therefore, as air pressure in an area increases, the density of the gas particles in that area increases.

3 0
3 years ago
What process removes carbon dioxide from the ocean?
Katen [24]

Answer:

B. decay of dead marine organisms

Explanation:

When the temperature is low, carbon dioxide is captured by the oceans, and when the temperature is high, it is released by the oceans into the atmosphere. At sea, carbon dioxide feeds phytoplankton.

Most of the carbon dioxide consumed by plant plankton (phytoplankton) returns to the atmosphere when this phytoplankton dies or is consumed, but a portion is deposited in the ocean floor sediments when these small particles sink. This process is called a "biological bomb" because carbon dioxide is transported from the atmosphere to the ocean floor.

8 0
3 years ago
A gas occupies a volume of 85.0 liters at a pressure of 2.24 atm and a temperature of 22.5 degrees celsius. How many moles of ga
White raven [17]

Answer:

n = 7.86 mol

Explanation:

This question can be solved using the ideal gas law of PV = nRT.

Temperature must be in K, so we will convert 22.5C to 295 K ( Kelvin = C + 273).

R is the ideal gas constant of 0.0821.

(2.24atm)(85.0L) = n(0.0821)(295K)

Isolate n to get:

n = (2.24atm)(85.0L)/(0.0821)(295K)

n = 7.86 mol

8 0
3 years ago
Calculate hydrochloric acid (umol)in 200 ul of a<br> 0.5173Msolution of acid?
strojnjashka [21]

<u>Answer:</u> The moles of hydrochloric acid is 1.0346\times 10^{-4}\mu mol

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Or,

\text{Molarity of the solution}=\frac{\text{Micro moles of solute}\times 10^6}{\text{Volume of solution (in }\mu L)}}

We are given:

Molarity of solution = 0.5173 M

Volume of solution = 200\mu L

Putting values in above equation, we get:

0.5173M=\frac{\text{Micro moles of HCl}\times 10^6}{200\mu L}\\\\\text{Micro moles of HCl}=1.0346\times 10^{-4}\mu mol

Hence, the moles of hydrochloric acid is 1.0346\times 10^{-4}\mu mol

7 0
3 years ago
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