Answer:
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Explanation:
Step 1: Data given
Mass of sodium bicarbonate = 2.7 grams
Step 2: The balanced equation
HCl + NaHCO3 ⇔ NaCl + H2O + CO2
Step 3: Calculate moles NaHCO3
moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles
Step 4: Calculate moles HCl
For 1 mol NaHCO3 we need 1 mol HCl
For 0.032 moles NaHCO3 = 0.032 moles HCl
Step 5: Calculate mass HCl
Mass HCl = moles HCl * molar mass HCl
mass HCl = 0.032 * 36.46 g/mol= 1.17 grams
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Answer:
66m
Explanation:
To get the area of something you multiple the length (5.5) by the width (12) together. So the problem would look like 5.5×12 and if you multipe that you get 66
Answer:
Explanation:
Flame test:
The metals ions can be detected through the flame test. Different ions gives different colors when heated on flame. Tom perform the flame test following steps should follow:
1. Dip a wire loop in the solution of compound which is going to be tested.
2. After dipping put the loop of wire on bunsen burner flame.
3. Observe the color of flame.
4. Record the flame color produce by compound
Color produce by metals:
Red = Lithium, zirconium, strontium, mercury, Rubidium (red violet)
Orange-red = calcium
Yellow = sodium, iron (brownish yellow)
Green = green
Blue = cesium. arsenic, copper, tantalum, indium, lead
Violet = potassium (lilac)
Answer:Ionic compounds form when positive and negative ions share electrons and form an ionic bond. ... The positive ion, called a cation, is listed first in an ionic compound formula, followed by the negative ion, called an anion. A balanced formula has a neutral electrical charge or net charge of zero.
Explanation:Simple ions:
Perchlorate ClO4- IO3-
Chlorate ClO3- BrO3-
Chlorite ClO2-
Hypochlorite OCl- OBr-
Answer:
1. 8.7moles of H2
2. 2.25moles of O2
Explanation:
1. 2NH3 —> N2 + 3H2
From the equation,
2moles of NH3 produce 3 moles of H2.
Therefore, 5.8moles of NH3 will produce Xmol of H2 i.e
Xmol of H2 = (5.8x3)/2 = 8.7moles
2. C3H8 + 5O2 —> 3CO2 + 4H2O
From the equation,
5moles of O2 produced 4moles of H2O.
Therefore, Xmol of O2 will produce 1.8mol of H2O i.e
Xmol of O2 = (5x1.8)/4 = 2.25moles