With what volume of 5.00 m hf will 4.72 g of calcium hydroxide react completely, according to the following reaction? 2hf + ca(oh)2 → caf2 + 2h2o

 The volume  of 5.00 M  HF that will react with 4.72  g   of  calcium hydroxide  is 0.0256 L

  calculation

2HF   + Ca(OH)₂ →  CaF₂ + 2H₂O

Step 1:  find the  moles  of Ca(OH)₂

moles= mass÷  molar mass

from periodic  table the molar mass of Ca(OH)₂  = 40 +( 16+1)₂= 74 g/mol

moles =  4.72 g÷  74 g/mol =0.0638   moles

Step  2:  use the  moles ratio to determine the  moles of Hf

from the equation above HF: Ca(OH)₂  is 2:1

therefore the moles of HF = 0.0638 moles x 2/1 =0.1276 moles

Step 3: find   the  volume  of HF

volume=moles /molarity

molarity =5.00 M = 5.00 mol/L

=0.1276  moles / 5.00 mol/l =0.0256 L

answer; i believe the correct answer is (c) /// is a homogeneous mixture of gaseous substance nitrogen,oxygen and smaller amounts of other substancesmixtures can be either ( homogeneous or heterogeneous)plz check first good luck

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