The number of years that must be invested at a rate of 7 % to earn$ 303.52 in interest is
8 years
<em><u>calculation</u></em>
- <em><u> </u></em><em>by use of the formula A = P (1+ rt) </em>
- <em> where : A is the final amount = 542 + 303.52 =$ 845.52</em>
<em> P is the principal money to be invested = $ 542</em>
<em> r= rate= 7/100=0.07</em>
<em> t= time required</em>
<em>=$ 845.52=$ 542( 1+ 0.07 t)</em>
- <em>open the bracket</em>
- <em>= $845.53= $542 + $37.94 t</em>
- <em>like terms together</em>
=$ 845.53 -$542 = $37.94 t
=$303.52 =$37.94 t
- divide both side by $37.94
= $303.52/ $ 37.94 = $37.94t/$37.94
t= 8 years
217 ml of 0.05 M HCl will be required to react with 5 gm of MnO2 according to the equation given.
Explanation:
The balanced chemical equation is given as:
4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g)
This shows that 4 moles of HCl reacts with 1 mole of MnO2
the mass of Manganese oxide is given as 5 grams
molar mass of MnO2 = 86.93 grams/mole
number of moles of MnO2 is given by
number of moles = 
number of moles= 
= 0.0575 moles of MnO2
From the equation:
4 moles of HCl reacts with 1 mole of MnO2
x moles of HCl reacts with 
moles of MnO2
=
= 0.23 moles of HCl will react
atomic mass of HCl = 36.46 Grams/mole
mass = 0.23 x 36.46
= 8.3858 grams.of HCl
molarity of HCl = 
volume is 1 litre
so molarity is 0.23 M
Using the formula
M1V1 = M2V2
0.05 x 1 = 0.23 X x
x = 0.217 litre
so 217 ml of 0.05 M HCl will be required to react with 5 gm of MnO2.
The input for cellular response is O2
Answer: Option A) 2 atoms
Explanation:
4Al + 302 ---> 2Al2O3
4 moles of Al reacts with 3 molecules of O2
So 1.10 moles of Al should reacts with Z molecules of O2 (Assume Z is unknown)
i.e 4 moles of Al = 3 molecules of O2
4 moles of Al = 6 atoms of O2 (since 1 molecule of oxygen has two atoms of oxygen)
1.10 atoms of Al = Z atoms of O2
To get the value of Z, cross multiply
Z x 4 moles = 6 atoms x 1.10 moles
4Z = 6.6
Divide both sides by 4
4Z/4 = 6.6/4
Z = 1.65 atoms (approximately 2 atoms)
Thus, 2 atoms of the excess reactant oxygen remains.
Answer:
C₂H₃O
Explanation:
From the question given above, the following data were obtained:
Carbon (C) = 48.8383%
Hydrogen (H) = 8.1636%
Oxygen (O) = 43.1981%
Empirical formula =?
The empirical formula of the compound can be obtained as follow:
C = 48.8383%
H = 8.1636%
O = 43.1981%
Divide by their molar mass
C = 48.8383 / 12 = 4.07
H = 8.1636 / 1 = 8.1636
O = 43.1981 / 16 = 2.7
Divide by the smallest
C = 4.07 / 2.7 = 2
H = 8.1636 / 2.7 = 3
O = 2.7 /2.7 = 1
Thus, the empirical formula of the compound is C₂H₃O
