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DanielleElmas [232]

3 years ago

8

The combustion reaction of propane is described by the reaction. C3H8 + 5O2 ? 4H2O + 3CO2. How many moles of O2 are required to

generate 3 moles of CO2?

1 answer:

Minchanka [31]3 years ago

7 0

the Answer would be one mole

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I WILL GIVE 15 POINTS PLS I NEED THIS BEFORE THE CLASS FINISHES IN 30 MINN

bulgar [2K]

Answer:

likely be the same

Explanation:

this is because we have one color that both atoms share (green). both sample 1 and sample 2 have green and another color. yet, since they share one color, they are likely similar

7 0

3 years ago

A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre

givi [52]

Answer:

$\mathbf{0.02 M}$

Explanation:

$\text{So, from the given question:}$

$\text{EDTA will make complex with}$ $V^{+3}$ $\text{and the remaining EDTA will react with }$ $Ga^{+3}$

$\text{Hence, the total concentration of}$ $V^{+3}$ & $Ga^{+3}$ $\text{will be equivalent to EDTA concentration.}$

$V_{EDTA} = 25 \ mL$

$V_{V^{+3}} = 59.0 \ mL$

$V_{Ga^{+3}} = 13.0 \ mL$

$M_{EDTA} = 0.0680 \ M$

$M_{V^{+3}} = ???(unknown)$

$M_{Ga^{+3}} = 0.0400 \ M$

$V^{+3} + EDTA \to V[EDTA] + EDTA(Excess) \to^{CoA} \ Ga[EDTA] _{complex}$

$M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})$

$0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18$

$x = \dfrac{1.18}{59}$

$\mathbf{x =0.02 \ M }$

5 0

3 years ago

What is oxidation state?

solniwko [45]

Answer:

a number assigned to an element in chemical combination which represents the number of electrons lost (or gained, if the number is negative), by an atom of that element in the compound

Explanation:

sodium, magnesium, iron is zero.

4 0

3 years ago

Read 2 more answers

If an atom has six electrons how many protons does it have

Sergio039 [100]

6 protons it have to be neutral

7 0

3 years ago

How many grams are in 4.07x10^15 molecules of calcium hydroxide

maksim [4K]

There are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by multiplying the number of moles in the substance by its molecular mass.

However, given that the number of molecules in calcium hydroxide is 4.07 x 10¹⁵molecules, we need to calculate the number of moles in Ca(OH)2 as follows:

no. of moles of Ca(OH)2 = 4.07 x 10¹⁵ ÷ 6.02 × 10²³

no. of moles = 0.676 × 10-⁸

no. of moles = 6.76 × 10-⁹ moles.

Molar mass of Ca(OH)2 = 74.093 g/mol

Mass of Ca(OH)2 = 74.093 × 6.76 × 10-⁹ moles

Mass = 5.01 × 10-⁷grams.

Therefore, there are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.

Learn more about how to calculate mass at: brainly.com/question/8101390?referrer=searchResults

5 0

2 years ago