Answer:

Explanation:
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In this case, for the given reactants side, we infer this is a double replacement reaction because all the cations and anions are switched around as a result of the chemical change, we infer that the products side include aluminum with nitrate and magnesium with sulfate as shown below:

However, we need to balance since unequal number of atoms are present at both sides, thus, we do that as shown below:

Thus, we make 6 Al atoms, 3 S atoms, 3 Mg atoms and 30 O atoms on each side in agreement with the law of conservation of mass.
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Answer:
Empirical formula is CaSO₄.
Explanation:
Given data:
Percentage of calcium =29.44%
Percentage of sulfur = 23.55%
Percentage of oxygen = 47.01%
Empirical formula = ?
Solution:
Number of gram atoms of Ca = 29.44 / 40 = 0.74
Number of gram atoms of S = 23.55 / 32 = 0.74
Number of gram atoms of O = 47.01 / 16 = 3
Atomic ratio:
Ca : S : O
0.74/0.74 : 0.74/0.74 : 3/0.74
1 : 1 : 4
Ca : S : O = 1 : 1 : 4
Empirical formula is CaSO₄.
Answer:
<em>20 Liters</em>
Explanation:
If the pressure is supposed to be constant, one of Charles - Gay Lussac's laws can be used to solve the exercise. His statement says that "the volume of the gas is directly proportional to its temperature at constant pressure", mathematically it would be:

Considering T₁ = 50 ° C; V₁ = 10.0 L; and T₂ = 100 ° C, we can calculate the volume V₂ Clearing it from the equation and replacing the values to perform the calculation:
V2= (V1 x T2) / T1 = (10.0 L x 100°C) / 50 °C = 20 L
Therefore, <em>the gas at 100 ° C will occupy a volume of 20.0 L</em>.
Answer:

Explanation:
The formula for efficiency is

Data:
Useful energy = 3 J
Energy input = 30 J
Calculation:

Answer:
237.2 mL.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
(XMV) acid = (XMV) base.
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HCl = (XMV) NaOH.</em>
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For HCl; X = 1, M = 0.5 M, V = ??? mL.
For NaOH, X = 1, M = 1.54 M, V = 77.0 mL.
<em>∴ V of HCl = (XMV) NaOH / (XV) HCl = (</em>1)(1.54 M)(77.0 mL) / (1)(0.5 M) = <em>237.2 mL.</em>