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Svetlanka [38]
3 years ago
6

The volume of oxygen required for complete combustionof a mixture of 5 cm3 of ch4 and5 cm3 c2h4

Chemistry
1 answer:
Stels [109]3 years ago
5 0
3 to 9 is the volume requires to complete. Ombustion
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Kobe is testing whether 100 g of substance A, 100 g of substance B, or 100 g of substance C produces more oxygen in a chemical r
skad [1K]
C, the amount of oxygen produced is the dependent variable since that is what is being measured and it is dependent on which substance is being tested
4 0
3 years ago
Numerator and denominator are Equal to one another? <br> I don’t get the concept
Alekssandra [29.7K]

It's like saying that if 1 meter is equal to 100 cm,

\frac{100cm}{1m}

or

\frac{1m}{100cm}

You can base the answers from the chart

5 0
3 years ago
What is the hydrogen ion concentration, [H+], of a solution with a pH of 5.43?
grigory [225]

Answer: 3.7 x10−6 Mole per dm^3

Explanation:

pH is the negative logarithm of hydrogen ion concentration in a solution.

So, pH = - log(H+)

Since the solution has a pH of 5.43

5.43 = -log(H+)

To get hydrogen ion concentration, find the Antilog of 5.43

(H+) = Antilog (-5.43)

(H+) = 0.000003715

Then, 0.000003715 in standard form becomes 3.7 x10−6 M

Thus, the concentration of hydrogen ion in the solution is 3.7 x10−6 Mole per dm^3

6 0
3 years ago
What is the percent composition in chloric acid (HClO3)?
Nuetrik [128]
Total weight = 1 + 35.45 + 3 * 16 = 84.45

H = 1 / 84.45 * 100% = 1.18%
Cl = 35.45 / 84.45 * 100% = 41.98%
O = 48 / 84.45 * 100% = 56.84%
7 0
3 years ago
How many grams of phosphorus are in 500.0 grams of calcium phosphide? (i need the work also)
kvv77 [185]

Answer:

\boxed{\text{170.0 g P}} 

Explanation:

The formula of calcium nitride is Ca₃P₂.

The masses of each element are:

\begin{array}{lrcr}\text{3Ca:} & 3 \times 40.08&=& \text{120.24 u}\\\text{2P:} & 2\times 30.97&=& \text{61.94 u}\\& \text{TOTAL} & = & \text{182.18 u}\\\end{array}

So, there are 61.94 g of P in 182.18 g of Ca₃P₂.

In 500 g of Ca₃P₂:

\text{Mass of P} = \text{500.0 g Ca$_{3}$P$_{2}$} \times \dfrac{\text{61.94 g P}}{\text{182.18 g Ca$_{3}$P$_{2}$}} = \text{170.0 g P}

There are \boxed{\textbf{170.0 g P}} in 500.0 g of Ca₃P₂.

3 0
3 years ago
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