If the mass of a sample of Mg is 0.500 g and is placed in 100.0 g of HCl, the heat gained is 546.5 J. What is the change in temperature assuming the specific heat of HCl is 4.017 J/K*g?

1.36 K

Explanation:

We have the following data:

The heat gained by HCl is calculated as:

heat = m x Sh x ΔT

where m is the mass of HCl (100.0 g), Sh is the specific heat of HCl (4.017 J/K.g) and ΔT is the change in temperature. We introduce the data in the mathematical expression to calculate ΔT as follows:

ΔT = heat/(m x Sh) = 546.5 J/(100.0 g x 4.017 J/K.g) = 1.36 K

Therefore, the change in temperature is 1.36 K.

periodic trends is an atom that gets larger as the number of electronic shells increase, the radius of atoms increases as you go down a certain group in the periodic table of elements. in general, the size of an atom will decrease as you move from left to right of a certain period.

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