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wlad13 [49]
3 years ago
15

HELP . Read the following graduated cylinder.

Chemistry
1 answer:
Anon25 [30]3 years ago
3 0

The volume of the liquid from the graduated cylinder is 22.5 mL

The last option (22.5 mL) is the correct option

The question seems to be incomplete

The image that completes the question is shown in the attachment below.

From the image, we observe that the liquid meniscus is between the 20 mL and 25 mL marks. That means the volume will lie in this range.

NOTE: The meniscus of a liquid is the upward or downward curve seen at the top of a liquid in a container.

The type of curve shown in the image is a concave meniscus and the volume of the liquid is read from the bottom of the curve.

To read the value from the graduated cylinder, we will first determine the difference between two consecutive lines. Between the 20 mL and 25 mL mark (two thick lines), we have 5 spaces separated by thin lines.

The measure between two consecutive lines is the difference between two thick lines divided by number of spaces between them

∴ The measure between two consecutive lines = (25mL - 20mL) ÷ 5

The measure between two consecutive lines = 5mL ÷ 5 = 1 mL

Now, to read the position of the liquid,

Starting from the 20 mL mark, the liquid exceeds two more lines + half way to the next line ( that is 2mL + 0.5 mL)

∴ The position of the liquid is 20 mL + 2mL + 0.5 mL = 22.5 mL

Hence, the volume of the liquid from the graduated cylinder is 22.5 mL

Learn more here: brainly.com/question/19240870

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n the reaction Mg (s) + 2HCl (aq) H2 (g) + MgCl2 (aq), how many moles of hydrogen gas will be produced from 75.0 milliliters of
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0.075 L * 1.0 M = 0.075 mol HCl
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What is the balanced NET ionic equation for the reaction when aqueous Cs₃PO₄ and aqueous AgNO₃ are mixed in solution to form sol
rusak2 [61]

Answer:

PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)

Explanation:

Hello!

In this case, since the net ionic equation of a chemical reaction shows up the ionic species that result from the simplification of the spectator ions, which are those at both reactants and products sides, we take into account that aqueous species ionize into ions whereas liquid, solid and gas species remain unionized. In such a way, for the reaction of cesium phosphate and silver nitrate we can write the complete molecular equation:

Cs_3PO_4(aq)+3AgNO_3(aq)\rightarrow Ag_3PO_4(s)+3CsNO_3(aq)

Whereas the three aqueous salts are ionized in order to write the following complete ionic equation:

3Cs^+(aq)+PO_4^{3-}(aq)+3Ag^+(aq)+3NO_3^-(aq)\rightarrow Ag_3PO_4(s)+3Cs^+(aq)+3NO_3^-(aq)

In such a way, since the cesium and nitrate ions are the spectator ions because of the aforementioned, the net ionic equation turns out:

PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)

Best regards!

7 0
3 years ago
A flask with a volume of 250.0 mL contains air with a density of 1.164 g/L. What is the mass of the air contained in the flask?
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Answer:

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2 years ago
Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con
arsen [322]

Answer:

a)  T_b=590.775k

b)  W_t=1.08*10^4J

d)  Q=3.778*10^4J

d)  \triangle V=4.058*10^4J

Explanation:

From the question we are told that:

Moles of N2 n=2.50

Atmospheric pressure P=100atm

Temperature t=20 \textdegree C

                      t = 20+273

                     t = 293k

Initial heat Q=1.36 * 10^4 J

a)

Generally the equation for change in temperature is mathematically given by

\triangle T=\frac{Q}{N*C_v}

  Where

  C_v=Heat\ Capacity \approx 20.76 J/mol/K

T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }

T_b-293k=297.775

T_b=590.775k

b)

Generally the equation for ideal gas is mathematically given by

 PV=nRT

For v double

 T_c=2*590.775k

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Therefore

PV=Wbc

Wbc=(2.20)(8.314)(1181_590.778)

Wbc=10805.7J

Total Work-done W_t

W_t=Wab+Wbc

W_t=0+1.08*10^4

W_t=1.08*10^4J

c)

Generally the equation for amount of heat added is mathematically given by

Q=nC_p\triangle T

Q=2.20*2907*(1181.55-590.775)\\

Q=3.778*10^4J

d)

Generally the equation for change in internal energy of the gas is mathematically given by

\triangle V=nC_v \triangle T

\triangle V=2.20*20.76*(1181.55-293)k

\triangle V=4.058*10^4J

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3 years ago
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