A2.17 gm sample barium reacted completely with water what is the equation for the reaction how many milliliters of dry h2 evoled at 21degree celcius and 748mm of hg ?

400 mL

Explanation:

Given data:

Mass of barium = 2.17 g

Pressure = 748 mmHg (748/760 = 0.98 atm)

Temperature = 21 °C ( 273+ 21 = 294k)

Milliliters of H₂ evolved = ?

Solution:

chemical equation:

Ba + 2H₂O →  Ba(OH)₂ + H₂

Number of moles of barium:

Number of moles = mass/ molar mass

Number of moles = 2.17 g / 137.327 g/mol

Number of moles = 0.016 mol

Now we  will compare the moles of barium with H₂.

                       Ba        :       H₂

                         1         :         1

                  0.016        :     0.016

Milliliters of H₂:

PV = nRT

V = nRT/P

V = 0.016 mol ×  0.0821 atm. mol⁻¹.k⁻¹.L×294 k/0.98 atm

V = 0.39 atm. L/0.98 atm

V = 0.4 L

L to mL

0.4 × 1000 = 400 mL

the answer is a. co2, this would be colder than the : )

explanation:

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol