Sucrose, a sweet, white crystalline substance, C12 H22 O11, OBTAINED CHIEFLY FROM THE JUICE OF THE SUGAR CANE AND SUGAR BEET, BUT ALSO PRESENT IN SORGHUM, THE sugar maple, some palms, and various other plants, and having extensive nutritional, pharmaceutical, and industrial uses; any of the class of carbohydrates to which this substance belongs, as glucose, levulose, and lactose.
Answer:
The solution is basic.
Explanation:
We can determine the nature of the solution via determining which has the large no. of millimoles (acid or base):
- If no. of millimoles of acid > that of base; the solution is acidic.
- If no. of millimoles of acid = that of base; the solution is neutral.
- If no. of millimoles of acid < that of base; the solution is basic.
- We need to calculate the no. of millimoles of acid and base:
no. of millimoles of acid (HNO₃) = MV = (1.3 M)(75.0 mL) = 97.5 mmol.
no. of millimoles of base (NaOH) = MV = (6.5 M)(150.0 mL) = 975.0 mmol.
<em>∴ The no. of millimoles of base (NaOH) is larger by 10 times than the acid (HNO₃).</em>
<em>So, the solution is: basic.</em>
Answer:
54g of water
Explanation:
Based on the reaction, 1 mole of methane produce 2 moles of water.
To solve this question we must find the molar mass of methane in order to find the moles of methane added. With the moles of methane and the chemical equation we can find the moles of water produced and its mass:
<em>Molar mass CH₄:</em>
1C = 12g/mol*1
4H = 1g/mol*4
12g/mol + 4g/mol = 16g/mol
<em>Moles methane: </em>
24g CH₄ * (1mol / 16g) = 1.5 moles methane
<em>Moles water:</em>
1.5moles CH₄ * (2mol H₂O / 1mol CH₄) = 3.0moles H₂O
<em>Molar mass water:</em>
2H = 1g/mol*2
1O = 16g/mol*1
2g/mol + 16g/mol = 18g/mol
<em>Mass water:</em>
3.0moles H₂O * (18g / mol) =
<h3>54g of water</h3>
Answer:
D. Element.
Explanation:
Atoms form elements. Elements form molecules. Molecules form compounds.
Answer:
10.5g
Explanation:
First, let us calculate the number of mole of NaHCO3 present in the solution. This is illustrated below:
Volume = 250mL = 250/1000 = 0.25L
Molarity = 0.5M
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 0.5 x 0.25
Mole = 0.125 mole
Now, we shall be converting 0.125 mole of NaHCO3 to grams to obtain the desired result. This can be achieved by doing the following:
Molar Mass of NaHCO3 = 23 + 1 + 12 +(16x3) = 23 + 1 +12 +48 = 84g/mol
Number of mole of NaHCO3 = 0.125 mole
Mass of NaHCO3 =?
Mass = number of mole x molar Mass
Mass of NaHCO3 = 0.125 x 84
Mass of NaHCO3 = 10.5g
Therefore, 10.5g of NaHCO3 is needed.
