First situation:
IV: soda, gatorade, orange juice, and water
DV: state of IV listed above
Control: freezer, and ice tray
Second Situation:
IV: laundry detergent, water
DV: result of the squares after being washed
Control: chocolate, type of cloth, squares of cloth
Third Situation:
IV: Water used, pea plant
DV: growth of pea plant
Control: pots and amount of water plant gets each day
Answer: -
6
Explanation: -
The given unbalanced chemical equation is As + NaOH -- > Na3AsO3 + H2
We see there 3 sodium on the right side from Na3AsO3.
But there are only 1 sodium on the left from NaOH.
So we multiply NaOH by 3.
As + 3 NaOH -- > Na3AsO3 + H2
Now we see the number of Hydrogen on the left is 3.
But the number of hydrogens is 2 on the left.
So, we multiply to get both sides 6 hydrogen.
As + 6NaOH -- > Na3AsO3 + 3 H2
Rebalancing for Na,
As + 6NaOH -- > 2Na3AsO3 + 3 H2.
Finally balancing As,
2 As + 6 NaOH -- > 2Na3AsO3 + 3H2
The coefficient of the NaOH molecule in the balanced reaction is thus 6
Answer:
If fixed amount of gas is heated then the volume will increase because the heat will cause the molecules of gas to move freely and increase the kinetic energy.
Explanation:
If fixed amount of gas is heated then the volume will increase because the heat will cause the molecules of gas to move freely and increase the kinetic energy.
According to the Charle's law
The volume of given amount of gas is directly proportional to the temperature at constant pressure and number of moles of gas.
Mathematical expression;
V ∝ T
V = kT
V/T = k
if volume is changed from V1 to V2 and temperature change from T1 to T2 then,
V1/T1 = k V2/T2= k
V1/T1 = V2/T2
Answer:
Molecular formula for the gas is: C₄H₁₀
Explanation:
Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g
At STP → 1 atm and 273.15K
1 atm . 0.0336 L = n . 0.082 . 273.15 K
n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)
n = 1.500 × 10⁻³ moles
Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m
Now we propose rules of three:
If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C
58 g of gas (1mol) would have:
(58 g . 0.480) / 0.580 = 48 g of C
(58 g . 0.100) / 0.580 = 10 g of H
48 g of C / 12 g/mol = 4 mol
10 g of H / 1g/mol = 10 moles
