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3 years ago

The following reaction is done at

Chemistry

2 answers:

sergejj [24]3 years ago

4 0

Answer : The volume of $H_$ gas produced is, 305.8 L

Explanation :

First we have to calculate the moles of Ca.

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

Molar mass of Ca = 40 g/mol

$\text{Moles of Ca}=\frac{500.0g}{40g/mol}=12.5mol$

Now we have to calculate the moles of $H_2$ gas.

The balanced chemical reaction is:

$Ca(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2(g)$

From the balanced chemical reaction we conclude that,

As, 1 mole of Ca react to give 1 mole of $H_2$ gas

So, 12.5 mole of Ca react to give 12.5 mole of $H_2$ gas

Now we have to calculate the volume of $H_$ gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $H_2$ gas = 1.0 atm

V = Volume of $H_2$ gas = ?

n = number of moles $H_2$ = 12.5 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $H_2$ gas = $25^oC=273+25=298K$

Putting values in above equation, we get:

$1.0atm\times V=12.5mole\times (0.0821L.atm/mol.K)\times 298K$

$V=305.8L$

Thus, the volume of $H_$ gas produced is, 305.8 L

bagirrra123 [75]3 years ago

3 0

Answer: 280dm3

Explanation:Please see attachment for explanation

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2 years ago

How many milliliters of sterile water for injection should be added to a vial containing 5 mg/mL of a drug to prepare a solution

jonny [76]

Answer:

7mL of sterile water is the initial amount of the concentrated solution is 3mL

Explanation:

In this problem, the vial must be <em>diluted </em>from 5mg/mL to 1.5mg/mL, that means the solution must be diluted:

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3 years ago

Lactic acid results from which of the following?

vazorg [7]

Explanation:

Lactic acid is formed during the breakdown of glucose. This is sometimes called "blood sugar."

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3 years ago

Read 2 more answers

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:

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

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3 years ago

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Wewaii [24]

Answer:

Explanation:

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3 years ago