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-Dominant- [34]

3 years ago

15

Who formulated the theory of evolution. lemark newton darwin lavisor

2 answers:

svet-max [94.6K]3 years ago

6 0

Answer:

(LeMark) formulated the theory of evolution.

Explanation:

Lubov Fominskaja [6]3 years ago

4 0

Lamarck was the first person to propose a fully formed theory of evolution.

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If the pH of vinegar is 3.0, what is the concentration of H+ in vinegar?

Cerrena [4.2K]

Answer: The concentration of $H^{+}$ ions in vinegar is 0.001 M.

Explanation:

Given: pH = 3.0

pH is the negative logarithm of concentration of hydrogen ion.

The expression for pH is as follows.

$pH = - log [H^{+}]$

Substitute the value into above expression as follows.

$pH = - log [H^{+}]\\3.0 = - log [H^{+}]\\conc. of H^{+} = antilog (- 3.0)\\= 0.001 M$

Thus, we can conclude that the concentration of $H^{+}$ ions in vinegar is 0.001 M.

5 0

2 years ago

Timmy and his dad are throwing rocks into the pond. The rocks that Timmy's dad throws go farther and faster. Which of

7nadin3 [17]

Can you put a picture

3 0

2 years ago

Which explanations did you include in your response? Check all that apply. The oxidation number of carbon changes from -1 to +4.

Kaylis [27]

Answer:

B, E

Explanation:

Edge

5 0

2 years ago

Given the balanced equation, 2 H2 + O2 -> 2 H2O, answer the following

9966 [12]

Answer: $O_2$ is the limiting reagent

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} H_2=\frac{10.0g}{2g/mol}=5.00moles$

$\text{Moles of} O_2=\frac{20.0g}{32g/mol}=0.625moles$

The balanced chemical reaction is :

$2H_2+O_2\rightarrow 2H_2O$

According to stoichiometry :

1 moles of $O_2$ require = 2 moles of $H_2$

Thus 0.625 moles of $O_2$ will require= $\frac{2}{1}\times 0.625=1.25moles$ of $H_2$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent as it is present more than the required amount.

8 0

3 years ago

"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the

soldier1979 [14.2K]

<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of hot water = 46.7 g

$m_2$ = mass of cold water = 45.33 g

$T_{final}$ = final temperature = 59.4°C

$T_1$ = initial temperature of hot water = 80.6°C

$T_2$ = initial temperature of cold water = 40.6°C

$c_1$ = specific heat of hot water = 4.184 J/g°C

$c_2$ = specific heat of cold water = 4.184 J/g°C

$c_3$ = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

$46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)$

$c_3=30.68J/g^oC$

Hence, the specific heat of calorimeter is 30.68 J/g°C

6 0

3 years ago