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3 years ago

What are the lengths of the diagonals of the kite?

Chemistry

1 answer:

umka21 [38]3 years ago

8 0

The answer ( 13 and 8 )

x²=5²+12²

x²=25+144

x²=169

x=13

x²=5²+6²

x²=25+36

x²=61

x=7.8

x=8

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You are on an island and need freshwater to drink. Which process could you

stepladder [879]

C distillation you would evaporate the water into another container leaving you with the salt free water

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3 years ago

The accepted value is 29.35. Which correctly describes this student's experimental data?

kkurt [141]

Answer:

- Both accurate and precise.

Explanation:

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3 years ago

Please help me if possible<br>

Gwar [14]

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2 years ago

How to draw Hess' Cycle for this question ?

NISA [10]

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ/mole$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.7kJ/mole$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.9kJ/mole$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ/mole$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)$

$\Delta H=51.8kJ/mole$

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

7 0

3 years ago

Why do molecular compounds have low melting points and low boiling points relative to ionic substances?

kap26 [50]

Answer:

The melting and boiling points of molecular compounds are generally quite low compared to those of ionic compounds. This is because the energy required to disrupt the intermolecular forces between molecules is far less than the energy required to break the ionic bonds in a crystalline ionic compound

4 0

3 years ago