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tensa zangetsu [6.8K]
4 years ago
15

Points D and E are midpoints of the sides of triangle ABC. The perimeter of the triangle is 48 units. What is the value of t?

Chemistry
2 answers:
Debora [2.8K]4 years ago
5 0

The value of t is 2

the correct answer is A

weeeeeb [17]4 years ago
3 0
     If points E and D are midpoints of triangle ABC, then the perimeter of the triangle is:AB + BC + CA = 48, or:
3 t + 3 t + 4 t + 4 t + 7 t + 6 = 4821 t = 48 - 621 t = 42t = 42 : 21t = 2Answer: A.  2.

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Identify the balanced equation for the following reaction:<br><br> SO2(g) + O2(g) → SO3(g)
sergiy2304 [10]

Answer:  The balanced equation for the given reaction is

2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g).

Explanation:

A chemical equation which contains same number of atoms on both reactant and product side.

For example, SO_{2}(g) + O_{2}(g) \rightarrow SO_{3}(g)

Here, number of atoms on reactant side are as follows.

  • S = 1
  • O = 4

Number of atoms on product side are as follows.

  • S = 1
  • O = 3

To balance this equation, multiply SO_{2} by 2 on reactant side and multiply SO_{3} by 2. Hence, the equation will be re-written as follows.

2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)

Here, number of atoms on reactant side are as follows.

  • S = 2
  • O = 6

Number of atoms on product side are as follows.

  • S = 2
  • O = 6

Now, there are same number of atoms on both reactant and product side. So, this equation is balanced.

Thus, we can conclude that the balanced equation for the given reaction is 2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g).

3 0
3 years ago
If chlorine gas is bubbles through an aqueous solution of sodium iodide,the result is elemental iodine and aqueous sodium chlori
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The balanced reaction that describes the reaction of chlorine gas and sodium iodide to produce elemental iodine and sodium chloride in aqueous solution is expressed Cl2+2NaI= I2 + 2NaCl. This kind of reaction is called single replacement reaction where the anion, in this case, is only replaced
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During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final
kompoz [17]

Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

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Excess reactant = (Fe(NO₃)₃

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n (FeSCN⁺²) = 0.000236 * 4.63

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V(standard 2) = 4.63 + 5.17

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M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

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