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emmainna [20.7K]
3 years ago
5

How to balance an equation?

Physics
2 answers:
Advocard [28]3 years ago
8 0
Do you mean a chemical equation? If so you balance it by adding coefficients.  (The subscripts shouldn't change, nor the products or reactants)
Let's say you have (reactant side) H2 + O2 ----> H2O (product side)

To balance the equation count the atoms in products and reactants.
 There are 2 hydrogen atoms on both sides, so they are balanced, but the oxygen atoms aren't. To make it balanced, we can add a 2 in front of H2O 
H2 + O2-------> 2H2O (4 hydrogen atoms, 2 oxygen atoms).

Now we need to adjust the reactant side since there are 4 hydrogen on the product side. We can do this by adding a 4 in front of the H2: 
2H2+O2-----> 2H2O

This is a balanced eqation: reactant side: 4 hydrogen, 2 oxygen
                                                product side: 4 hydrogen, 2 oxygen 
vova2212 [387]3 years ago
3 0
To balance an equation, you do whatever you did on one side, on the other side. If you subtract, then you subtract on the other side. If you multiply, you multiply on the other side, etc...
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Shtirlitz [24]

Answer:

D)     D = \frac{5}{4}  - \frac{3}{4} \ C, E)  (C, D) = ( \frac{17}{7}, \ \frac{-4}{7}

Explanation:

Part D) two expressions are indicated

          3C + 4D = 5

          2C +5 D = 2

let's simplify each expression

         3C + 4D = 5

         4D = 5 - 3C

we divide by 4

            D = \frac{5}{4}  - \frac{3}{4} \ C

The other expression

       2C +5 D = 2

       2C = 2 - 5D

        C = 1 -  \frac{5}{2} \ D

we can see that the correct result is 1

Part E.

It is asked to solve the problem by the substitution method, we already have

          D =  \frac{5}{4}  - \frac{3}{4} \ C

we substitute in the other equation

            2C +5 D = 2

             2C +5 (5/4 - ¾ C) = 2

we solve

            C (2 - 15/4) + 25/4 = 2

             -7 / 4 C = 2 - 25/4

             -7 / 4 C = -17/4

              7C = 17

               C = \frac{17}{7}

now we calculate D

               D = \frac{5}{4} - \frac{3}{4} \ \frac{17}{7}

               D = 5/4 - 51/28

               D =\frac{35-51}{28}

               D = - 16/28

               D = - \frac{4}{7}

the result is (C, D) = ( \frac{17}{7}, \ \frac{-4}{7} )

8 0
3 years ago
The atoms ,molecules,or compound present at the start of a chemical reaction that parcitipate in the reaction . which part of a
Tomtit [17]

Answer:

The atoms ,molecules, or compound present at the start of a chemical reaction that parcitipate in the reaction are <u><em>the reactants.</em></u>

Explanation:

The chemical reaction is the way in which one substance reacts against another. So, a chemical reaction consists of the transformation of some substances into others, that is, the process of arranging atoms and bonds when chemical substances come into contact.

In a chemical reaction, the initial substances are called reactants, while the new substances obtained are called products.

So, <u><em>the atoms ,molecules, or compound present at the start of a chemical reaction that parcitipate in the reaction are the reactants.</em></u>

8 0
4 years ago
A storm cloud has an electric charge of (2.1400x10^1) C near the top of the cloud and (2.99x10^1) C near the bottom of the cloud
blagie [28]

Answer:

hey

Explanation:

4 0
3 years ago
How long does it take an automobile traveling 66.7 km/h to become even with a car that is traveling in another lane at 52.7 km/h
tresset_1 [31]

Answer:

The  time taken is  t =  32.5 \  s

Explanation:

From the question we are told that

   The  speed  of  first car is  v_1  =  66.7 \ km/h  =  18.3 \  m/s

    The  speed  of  second car is v_2  =  52.7 \ km/h  =  14.64 \  m/s

   The  initial distance of separation is  d =  119 \ m

The distance covered by first car is mathematically represented as

     d_t =  d_i  +  d_f

Here  d_i is the initial distance which is  0 m/s

  and  d_f  is the final distance covered which is  evaluated as d_f  =  v_1 * t

So

     d_t =  0 \  m/s  +  (v_1 * t )

     d_t =  0 \  m/s  +  (18.3 * t )

The distance covered by second  car is mathematically represented as

     d_t =  d_i  +  d_f

Here  d_i is the initial distance which is  119 m

  and  d_f  is the final distance covered which is  evaluated as d_f  =  v_2* t

       d_t =  119  + 14.64 *  t

Given that the two car are now in the same position we have that

    119  + 14.64 *  t  =   0   +  (18.3 * t )

   t =  32.5 \  s

6 0
3 years ago
CAN SOMEBODY PLEASE HELP ME! i need help and i wanna pass
umka21 [38]

Answer:it would be C

Explanation:

8 0
4 years ago
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