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3 years ago

How adaptations in a bird's beak help them survive in their environment

2 answers:

7 0

A adaptation in a birds beak helps them survive, because their beak adapted to the food source. Their beaks needs to be a certain size or shape to eat different types of food based on their living environment.

jeyben [28]3 years ago

3 0

Answer:

they can get food easily and they can pack the predators to help them with their safty!♥

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A_________is the distance from one compression to the next compression in a longitudinal wave. (Page 2 Vocabulary)

Marina86 [1]

Answer:

wavelength

Explanation:

4 0

3 years ago

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

3 years ago

What is the molality of the resulting solution when 15.0 g of Licl is combined with 105 g of water? 0/1 point 0 1.42 m 2.95 m 3.

xenn [34]

Answer:

3.37 m

Explanation:

<u>Number of moles of solute present in 1 kg of solvent is termed as molality</u>

It is represented by 'm'.

Thus,

$Molality\ (m)=\frac {Moles\ of\ the\ solute}{Mass\ of\ the\ solvent\ (kg)}$

Given that:

The mass of LiCl = 15.0 g

Molar mass of LiCl = 42.394 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{15.0\ g}{42.394\ g/mol}$

$Moles= 0.3538\ mol$

Mass of the solvent = 105 g

Also, 1 g = 0.001 g

So,

Mass of water (solvent) = 0.105 kg

Molality is:

$Molality\ (m)=\frac {0.3538}{0.105}\ m$

<u>Molality = 3.37 m</u>

3 0

3 years ago

URGENT! PLEASE HELP ASAP

barxatty [35]

14.7 is the right answer plus 86

4 0

3 years ago

If 4.4 moles of al react, how many moles of h2so4 are consumed?

astraxan [27]

2 Al + 3 H₂SO₄ = Al₂(SO₄)₃ + 3 H₂

2 moles Al -------- 3 moles H₂SO₄
4.4 moles Al ------ ?

moles H₂SO₄ = 4.4 x 3 / 2

moles H₂SO₄ = 13.2 / 2

= 6.6 moles of H₂SO₄

hope this helps!

4 0

3 years ago