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Sonja [21]
3 years ago
13

Can anyone draw a Lewis Structure for caprolactam (C6H11NO)

Chemistry
1 answer:
andreev551 [17]3 years ago
8 0
Answer: in pic
explanation:

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Rainwater can have a ph as low as 3,meaning that it is very
GREYUIT [131]

Rainwater can have a ph as low as 3, meaning that it is very C. Acidic

3 0
3 years ago
How many grams of phosphorus are in 500.0 grams of calcium phosphide? (i need the work also)
kvv77 [185]

Answer:

\boxed{\text{170.0 g P}} 

Explanation:

The formula of calcium nitride is Ca₃P₂.

The masses of each element are:

\begin{array}{lrcr}\text{3Ca:} & 3 \times 40.08&=& \text{120.24 u}\\\text{2P:} & 2\times 30.97&=& \text{61.94 u}\\& \text{TOTAL} & = & \text{182.18 u}\\\end{array}

So, there are 61.94 g of P in 182.18 g of Ca₃P₂.

In 500 g of Ca₃P₂:

\text{Mass of P} = \text{500.0 g Ca$_{3}$P$_{2}$} \times \dfrac{\text{61.94 g P}}{\text{182.18 g Ca$_{3}$P$_{2}$}} = \text{170.0 g P}

There are \boxed{\textbf{170.0 g P}} in 500.0 g of Ca₃P₂.

3 0
3 years ago
Scientist think that many species have one common ancestor. Which answer choice best tells why scientists think this way?
BigorU [14]

Answer:

homologous structures.

Explanation:

If you look at many different organisms, you will see similar structures. Like humans have a forearm, birds have a wing, whales have fins, leopards have legs. The bones evolvolved to fit the animal

3 0
3 years ago
Can someone please tell me what the answer is to all this
Citrus2011 [14]
Is that all you have...............................................................
8 0
3 years ago
A 10-gram sample of zinc loses 560 J of heat and has a final temperature of 100
marin [14]
<h3>Answer:</h3>

Initial temperature is 243.59°C

<h3>Explanation:</h3>

The quantity of heat is calculated by multiplying the mass of a substance by its specific heat capacity and change in temperature.

That is; Q = m×c×ΔT

In this case;

Quantity of heat = 560 J

Mass of the Sample of Zinc = 10 g

Final temperature = 100°C

We are required to determine the initial temperature;

This can be done by replacing the known variables in the formula of finding quantity of heat,

Specific heat capacity, c, of Zinc = 0.39 J/g.°C

Therefore,

560 J = 10 g × 0.39 J/g°C × ΔT

ΔT = 560 J ÷ (3.9 J/°C)

   = 143.59°C

But, since the sample of Zinc lost heat then the temperature change will have a negative value.

ΔT = -143.59°C

Then,

ΔT  = T(final) - T(initial)

Therefore,

T(initial) = T(final) - ΔT

              = 100°C - (-143.59°C)

              = 243.59°C

Hence, the initial temperature of zinc sample is 243.59°C

6 0
3 years ago
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