Answer : The volume of 6M NaOH stock solution is, 16.7 mL
Explanation :
To calculate the volume of NaOH stock solution, we use the equation given by neutralization reaction:
where,
are the molarity and volume of NaOH stock solution.
are the molarity and volume of NaOH.
We are given:
Putting values in above equation, we get:
Thus, the volume of 6M NaOH stock solution is, 16.7 mL
Answer:
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Answer:
191 °C
Explanation:
We'll begin by calculating the number of mole in 88.0 g of Ne. This can be obtained as follow;
Mass of Ne = 88.0 g
Molar mass of Ne = 20 g/mol
Mole of Ne =?
Mole = mass /molar mass
Mole of Ne = 88 / 20
Mole of Ne = 4.4 moles
Next, we shall determine the temperature. This can be obtained as follow:
Mole of Ne = 4.4 moles
Pressure (P) = 350 KPa
Volume (V) = 48.5 L
Gas constant (R) = 8.314 KPa.L/Kmol
Temperature (T) =?
PV = nRT
350 × 48.5 = 4.4 × 8.314 × T
16975 = 36.5816 × T
Divide both side by 36.5816
T = 16975 / 36.5816
T = 464 K
Finally, we shall convert 464 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 464 K
T(°C) = 464 – 273
T(°C) = 191 °C
Thus, the temperature is 191 °C
Answer:
For the complete combustion of 10 g of glucose 11.52 g of oxygen is required.
Explanation:
Given data:
Mass of glucose = 10 g
Mass of oxygen required for complete combustion = ?
Solution:
Chemical equation:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
Now we will convert the given mass of glucose into number of moles.
Number of moles = mass/ molar mass
Number of moles = 10 g/ 180.156 g/mol
Number of moles = 0.06 mol
Now we will compare the moles of glucose with oxygen.
C₆H₁₂O₆ : O₂
1 : 6
0.06 : 6×0.06 = 0.36 mol
Mass of oxygen required:
Mass = number of moles × molar mass
Number of moles = 0.36 mol × 32 g/mol
Number of moles = 11.52 g
For the complete combustion of 10 g of glucose 11.52 g of oxygen is required.
ANSWER:
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