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Margaret [11]
2 years ago
9

Select the Independent, Dependent and Controlled Variables

Chemistry
2 answers:
Maru [420]2 years ago
6 0

Answer:

The independent variable is graphed on the x-axis. The dependent variable, which changes in response to the independent variable, is graphed on the y-axis. Controlled variables are usually not graphed because they should not change.

Explanation:

dalvyx [7]2 years ago
4 0

Answer:

The independent variable is graphed on the x-axis. The dependent variable, which changes in response to the independent variable, is graphed on the y-axis. Controlled variables are usually not graphed because they should not change.

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what happens if you fail a quarter in high school, does that credit not count anymore ? and do i have to retake the class if i p
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3 years ago
How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at a pressure of 12.4 atm and a tempe
polet [3.4K]
First, we need the no.of moles of O2 = mass/molar mass of O2
                                                             = 55 g / 32 g/mol
                                                             = 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2 
So we can get the no.of moles of H2O = 2 * moles of O2
                                                                  = 2 * 1.72 mol
                                                                  = 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT

when P = 12.4 atm  & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K

12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
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4 0
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Can selicone conduct electricity and heat
tatuchka [14]
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Read 2 more answers
Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have
Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

7 0
2 years ago
This element is found in group 1, period 7 and has the lowest mass in this period. *
liraira [26]

The answer is Francium.

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