if 19.1 mL of 1.26 M HBr is headed to 23.7g of CAC03 what volume of CO2 would be produced as 713.6 torr and 29.9°C

0.32 L

Explanation:

The equation of the reaction is;

2HBr(aq) + CaCO3(s) >CaBr2(aq) + H2O(l) + CO2(g)

The number of moles HBr reacted = 19.1/1000 L * 1.26 M = 0.024 moles

According to the reaction equation;

2 moles of HBr produces 1 mole of CO2

0.024 moles of HBr produces  0.024 * 1/2 = 0.012 moles of CO2

For CaCO3

Number of moles  = 23.7g/100 g/mol = 0.237 moles

Since the reaction is 1:1, 0.237 moles of CO2 is produced.

Hence HBr is the limiting reactant and 0.012 moles of CO2 is produced.

From PV =nRT

V = nRT/P

P= 713.6 torr

n= 0.012 moles

T = 29.9°C + 273 = 302.9 K

R =  62.36 L torr/mol K

V = 0.012 * 62.36 * 302.9/713.6

V = 0.32 L

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.