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3 years ago

HELP QUICK!! Match the term with the definition. (4 points)

Chemistry

2 answers:

Rufina [12.5K]3 years ago

7 0

Answer:

1. c.

2. d.

3. b.

4. a.

Explanation:

Evgesh-ka [11]3 years ago

6 0

2. d

1. c

3. a

4. b

I am not sure since I don't know what plasma is

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Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm

8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

$\frac{dV}{dt}=10\frac{cm^3}{s}$

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

$\frac{dh}{dt} = ?*\frac{dV}{dt}$

Of course, $?=\frac{dh}{dV}$ .

Now, since the volume of a cone is $V=\pi r^2h/3$ and the ratio $r/h=15/20=3/4$ or $r=3/4h$ , the volume becomes:

$V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3$

We proceed to its differentiation:

$\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}$

Then, we compute $\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}$

Finally, at h=2:

$\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}$

Best regards.

4 0

3 years ago

A rubber balloon was filled with helium at 25.0˚C and placed in a beaker of liquid nitrogen at -196.0˚C. The volume of the cold

Ksenya-84 [330]

Answer:

The volume of helium at 25.0 °C is 60.3 cm³.

Explanation:

In order to work with ideal gases we need to consider absolute temperatures (Kelvin). To convert Celsius to Kelvin we use the following expression:

K = °C + 273.15

The initial and final temperatures are:

T₁ = 25.0 + 273.15 = 298.2 K

T₂ = -196.0 + 273.15 = 77.2 K

The volume at 77.2 K is V₂ = 15.6 cm³. To calculate V₁ in isobaric conditions we can use Charle's Law.

$\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}} \\V_{1}=\frac{V_{2}}{T_{2}} \times T_{1}=\frac{15.6cm^{3} }{77.2K} \times 298.2K=60.3cm^{3}$

3 0

3 years ago

2. What ions are present in what ratio in a solution of aqueous calcium chloride?

Alenkasestr [34]

Answer:

$\mathrm{Ca}^{2+} \text { and } \mathrm{Cl} \text { - ions are present in } 1: 2 \text { ratio in a solution of aqueous calcium chloride. }$

Explanation:

Here in Calcium Chloride ionic bond is present in between calcium and chlorine atoms. As we know according to Octet rule calcium have two excess atoms and for matching nearest noble gas electronic configuration. It donate two electrons to gain more stability and form $\mathrm{Ca}^{2+}$ , while chlorine is deficient from one electron to meet nearest noble gas electronic configuration therefore two chlorine atoms accept excess electron from calcium individually and form two $\mathrm{Cl}^{-}$ ions.

$\text { Equation is as follows: } \mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{CaCl}_{2}$

Hence aqueous solution of calcium chloride breaks the ionic bond pairing in one $\mathrm{Ca}^{2+}$ and two $\mathrm{Cl}^{-}$ ions: $\mathrm{CaCl}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \quad \mathrm{Ca}^{2+}(\mathrm{ag})+2 \mathrm{Cl}(\mathrm{ag})$