Fill in the blanks below. 1. The equation shows a the breaking and forming of chemical bonds that leads to a change in the composition of matter
2. In the equation, CO2 is a
3. In the equation, C6H1206 is a
4. In O2, the type of bond that holds the two oxygen atoms together is a
5. In H2O, the type of bond that holds one of the hydrogen atoms to the oxygen atom is a
6. The number of oxygen atoms on the left side of the equation is the number of oxygen atoms on the
Chemical equation
right side.
reactant
product
nonpolar
covalent bond
polar
covalent bond
equal to
different from
hydrogen bond

Explanation:

6CO₂ + 6 H₂O ⇄ C₆H₁₂0₆ + 6O₂

This is the chemical equation given .

1. The equation shows a __Chemical equationthe breaking and forming of chemical bonds that leads to a change in the composition of matter.

2. In the equation, CO₂ is a___reactant.

3. In the equation, C₆H₁₂0₆ is a ___product.

4. In O₂, the type of bond that holds the two oxygen atoms together is a_nonpolar_covalent bond.

5. In H₂O, the type of bond that holds one of the hydrogen atoms to the oxygen atom is a__polar_hydrogen bond.

6. The number of oxygen atoms on the left side of the equation is__equal to the number of oxygen atoms on the right side.

gas

explanation:

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              nitroswawdasdwasdawsdwasdwasdwasdwaswasdwasdawdswasdwadwadswadadadasd gay

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