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3 years ago

The following diagram of a pendulum shows the path it takes as it swings back and forth.

Chemistry

2 answers:

JulsSmile [24]3 years ago

6 0

The answer would be B

lbvjy [14]3 years ago

4 0

I believe it should be B

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Mars2501 [29]

Answer:

Explanation:

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8 0

2 years ago

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Dew forming on grass is an example of condensation true or fales and explain

exis [7]

True becuase dew is coming out of the air which if you look at a glass of water it has condensation on it becuase it is hot

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3 years ago

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2 attempts left Click the "draw structure" button to launch the drawing utility. Draw one product formed when the following dien

Yuliya22 [10]

Answer:

2-chloro-1-methyl-cyclohex-1,4-diene.

Explanation:

Hello,

In this case, the addition of hydrochloric acid acts as an electrophilic atack in which the hydrogen bonded to the double-bonded carbon connected to the carbon with the methyl substitution is substituted by the chlorine from the hydrochloric acid, in such a way, 2-chloro-1-methyl-cyclohex-1,4-diene is produced as one equivalent of HCl is used therefore one substitution will be attained for chlorine, and hydrogen as a side product as shown on the attached picture.

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7 0

2 years ago

The enthalpy of formation of MX is ΔHf° = –525 kJ/mol. The enthalpy of sublimation of M is ΔHsub = 139 kJ/mol. The ionization en

polet [3.4K]

I'm sorry but could you tell me IE2 I couldn't asks for details

3 0

2 years ago

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$

= (-3399.6) + (1560)

= -1839.6 KJ

5 0

3 years ago