I need help, please im currently failing :// Now that you have seen a few examples of chemical equations, get some practice balancing and categorizing chemical equations by providing some of your own. For this activity, think of two everyday chemical reactions. (You can select two of the ones you chose for your science journal response.)

Once you have chosen your two everyday chemical reactions, conduct research to determine the chemical equations that are used to represent those reactions. List the two reactions and their chemical equations in your post. Also, state the type of each reaction, such as synthesis, decomposition, combustion, or one of the displacement reactions. Then, explain why each reaction falls into that specific category.

After you post, examine at least two other students' chemical equations. Check to see if each equation follows the law of conservation of mass by trying to balance the equations. If the equations are balanced, state how you know they are correct. If the equations are not balanced, then state what the equations should be in order to be correct.

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.

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

honesty and integrity.

inspire others.

commitment and passion.

good communicator.

decision-making capabilities.

accountability.

delegation and empowerment.

creativity and innovation.

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