Answer:
If a gas has experienced a small increase in volume but has maintained the same pressure and number of moles, the temperature of the gas will DROP.
Explanation:
According to Boyle’s law of ideal gases, volume and temperature of a gas is inversely related, as long as the pressure is kept constant;
P₁V₁/T₁ = P₂V₂/T₂
Therefore, if the volume of the gas increases, the temperature will definitely decrease due to the inverse relationship. The gas will get cooler.
Learn More:
For more on Boyle's Law check out;
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Answer:
350mmHg
Explanation:
Use Dalton law
Total=P gas 1+p gas 2+ P gas 3
825=P1+350+125
825=P1+475
825-475= P1
P1= 350 mm Hg
Answer:
d) PO4^3-, HPO4^2-
Explanation:
Basically, an acid and a base which differs only by the presence or absence of proton (hydrogen ion) are called a conjugate acid-base pair.
a) HI, I
This is incorrect. For the acid, HI the conjugate base is I⁻ ion.
b) HCHO2, SO4^2-
This is incorrect, there's no relationship between both entities.
c) CO3^2-, HCI
This is incorrect, there's no relationship between both entities.
d) PO4^3-, HPO4^2-
This is correct. The difference between both entities is the Hydrogen ion. This is the conjugate acid-base pair
Answer:
11.45kcal/g
2.612 × 10³ kcal
Explanation:
When a compound burns (combustion) it produces carbon dioxide and water. The combustion of 2-methylheptane can be represented by the following balanced equation:
2 C₈H₁₈ + 25 O₂ ⇄ 16 CO₂ + 18 H₂O
It releases 1.306 × 10³ kcal every 1 mol of C₈H₁₈ that is burned.
<em>What is the heat of combustion for 2-methylheptane in kcal/gram?</em>
We know that the molar mass of C₈H₁₈ is 114.0g/mol. Then, using proportions:
<em>How much heat will be given off if molar quantities of 2-methylheptane react according to the following equation? 2 C₈H₁₈ + 25 O₂ ⇄ 16 CO₂ + 18 H₂O</em>
In this equation we have 2 moles of C₈H₁₈. So,
Answer:
The answer to your question is 21.45 g of KBr
Explanation:
Chemical reaction
2K + Br₂ ⇒ 2KBr
14.4 ?
Process
1.- Calculate the molecular mass of bromine and potassium bromide
Bromine = 2 x 79.9 = 159.8g
Potassium bromide = 2(79.9 + 39.1) = 238 g
2.- Solve it using proportions
159.8 g of Bromine ------------ 238 g of potassium bromide
14.4 g of Bromine ------------ x
x = (14.4 x 238) / 159.8
x = 3427.2 / 159.8
x = 21.45g of KBr
