Answer:
5746.0 mL.
Explanation:
We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁</em>
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V₁ = 6193.0 mL, T₁ = 62.3°C + 273 = 335.3 K.
V₂ = ??? mL, T₂ = 38.1°C + 273 = 311.1 K.
<em>∴ V₂ = V₁T₂/T₁ </em>= (6193.0 mL)(311.1 K)/(335.3 K) = <em>5746.0 mL.</em>
Answer:
1.78 × 10⁹ μg
Explanation:
We have to convert 1.78 kg to μg.
Step 1: Convert 1.78 kilograms to grams
We will use the conversion factor 1 kg = 10³ g.
1.78 kg × 10³ g/1 kg = 1.78 × 10³ g
Step 2: Convert 1.78 × 10³ grams to micrograms
We will use the conversion factor 1 g = 10⁶ μg.
1.78 × 10³ g × 10⁶ μg/1 g = 1.78 × 10⁹ μg
Answer:
double replacement is the answer
Answer:
The concentration of monosodium phosphate is 0.1262M
Explanation:
The buffer of H₂PO₄⁻ / HPO₄²⁻ (Monobasic phosphate and dibasic phosphate has a pKa of 7.2
To determine the pH you must use Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
<em>Where [A⁻] is molarity of the conjugate base of the weak acid, [HA].</em>
For H₂PO₄⁻ / HPO₄⁻ buffer:
pH = 7.2 + log [HPO₄⁻² ] / [H₂PO₄⁻]
As molarity of the dibasic phosphate is 0.2M and you want a pH of 7.4:
7.4 = 7.2 + log [0.2] / [H₂PO₄⁻]
0.2 = log [0.2] / [H₂PO₄⁻]
1.58489 = [0.2] / [H₂PO₄⁻]
[H₂PO₄⁻] = 0.1262M
<h3>The concentration of monosodium phosphate is 0.1262M</h3>
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Answer:
A. The energy of moving particles near a fire
Explanation:
