A student heats a 2.5 g sample of calcium carbonate in a test tube. What mass of calcium oxide remains in the test tube when the reaction is complete? CaCO3(s) = CaO(s) + CO2(g)

Select one:
a. 1.4 g CaO
b. 2.1 g CaO
c. 2.5 g CaO
d. 1.0 g CaO

1.4 g CaO

Explanation:

Given data:

Mass of calcium carbonate = 2.5 g

Mass of Cao remain in tube = ?

Solution:

Chemical equation:

CaCO₃    →     CaO + CO₂

Number of moles of CaO:

Number of moles = mass/molar mass

Number of moles = 2.5 g/ 100 g/mol

Number of moles = 0.025 mol

now we will compare the moles of CaO with CaCO₃.

                    CaCO₃        :          CaO

                        1              :             1

                    0.025         :         0.025

Mass of CaO:

Mass = number of moles × molar mass

Mass = 0.025 mol × 56 g/mol

Mass = 1.4 g

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

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