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irina [24]
3 years ago
5

A graduated cylinder contains 20.8 mL of water. What is the new water level, in milliliters, after 35.2 g of silver metal is sub

merged in the water?
Chemistry
1 answer:
stepladder [879]3 years ago
5 0

<u>Answer:</u> The new water level of the cylinder is 24.16 mL

<u>Explanation:</u>

To calculate the volume of water displaced by silver, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of silver = 10.49 g/mL

Mass of silver = 35.2 g

Putting values in above equation, we get:

10.49g/mL=\frac{35.2g}{\text{Volume of silver}}\\\\\text{Volume of silver}=\frac{35.2g}{10.49g/mL}=3.36mL

We are given:

Volume of graduated cylinder = 20.8 mL

New water level of the cylinder = Volume of graduated cylinder + Volume of water displaced by silver

New water level of the cylinder = (20.8 + 3.36) mL = 24.16 mL

Hence, the new water level of the cylinder is 24.16 mL

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The solubility of N2in blood at 37°C and at a partial pressure of 0.80 atm is 5.6 × 10−4mol/L. A deep-sea diver breathes compres
Luda [366]

Solution :

According to Henry's law of solubility, we have c = kp

Henry's law constant, $k=\frac{c}{p}$

                                      $=\frac{0.00056}{0.80}$

                                      = 0.0007 mol/L.atm

When the pressure is = 4 atm

The solubility is c = 0.0007 mol/(L.atm) x 4 atm

                             = 0.0028 mol/L

Therefore, in a 5 liter of blood, the moles of nitrogen dissolved

= 0.0028 x 5

= 0.014 moles

At the surface, the solubility is = 0.00056 mol/L

So the moles of the nitrogen dissolved = 5 x 0.00056

                                                                  = 0.0028 moles

Therefore, the number of moles of nitrogen released = 0.014 - 0.0028

                                                                                         = 0.0112 moles

Given total pressure = 1 atm                                                                        

Temperature = 37 degree C

                     = 37 + 273

                      = 310 K

R = $0.0821\ L -atm/ mol.K $

Therefore, volume of the nitrogen is

$V=\frac{nRT}{P}$

$V=\frac{0.0112 \times 0.0821 \times 310}{1}$

  = 0.285 L

5 0
2 years ago
State the postulates of Dalton's atomic theory. ​
horrorfan [7]
A theory of chemical combination, first stated by John Dalton in 1803. It involves the following postulates: (1) Elements consist of indivisible small particles (atoms). (2) All atoms of the same element are identical; different elements have different types of atom. (3) Atoms can neither be created nor destroyed.

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4 0
2 years ago
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The rate constant for the oxidation of nitric oxide by ozone is 2 x 10^14 molecule cm s, whereas that for the competing reaction
andreev551 [17]

Answer:

The NO + O3 is the dominant reaction.

Explanation:

First of all, let's convert to molecules/cm³;

For O3;

O3 at 40 ppb in atm= 4 x 10^(-8) atm and from ideal gas law PV = nRT or simplify n/V = P/RT

Thus, plugging in the relevant values to get;

n/V = [4 x 10^(-8)]/(0.0821 x 298) = 1.636 x 10^(-9)

So, n/V = 1.636 x 10^(-9) = (1.635 x 10-9 mol L-1)(6.02 x10^(23) molec/mol)(L/1000 cm3) =

9.84 x 10^(11) molecules/cm³

But from the question, NO has 2 moles, and thus concentration is;

2 x 9.84 x 10^(11) = 1.968 x 10^(12) molec/cm³

For O2;

Following the same pattern for O3, we obtain;

(0.21 atm)/[(0.0821 L atm mol-1 K-1)(298K)] = 5.167 x 1018 molecules/cm³

Now, for NO and O3 reaction the rate is; k[NO] [O3]

Thus rate;

= (2 x 10^(-14)cm³/molec.s)( 9.84 x 10^(11)molec/cm³)(1.968 x 10^(12) molec/cm³) = 3.9 molec/cm³.s

For 2NO + O2 → 2NO2 reaction, rate = k[NO]2 [O2]

Thus, rate;

= (2 x 10^(-38) cm^(6)/molec².s )( 1.968 x 10^(12) molec/cm³) ²

(5.167 x 1018 molec/cm³)

= 40,000 molec/cm³.s

Observing the two rates, it's clear that the NO + O3 is the dominant reaction.

6 0
2 years ago
Which of the following is fact-based science rather than part of a belief system
makkiz [27]

Answer:

c. chemistry

Explanation:

it uses science and math which are based on facts.

6 0
2 years ago
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A 60.0 g block of iron that has an initial temperature of 250. °C and 60.0 g bloc of gold that has an initial temperature of 45.
Maslowich

Answer:

The final temperature at the equilibrium is 204.6 °C

Explanation:

Step 1: Data given

Mass of iron = 60.0 grams

Initial temperature = 250 °C

Mass of gold = 60.0 grams

Initial temperature of gold = 45.0 °C

The specific heat capacity of iron = 0.449 J/g•°C

The specific heat capacity of gold = 0.128 J/g•°C.

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qiron = -Qgold

Q=m*c*ΔT

m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)

⇒with m(iron) = the mass of iron = 60.0 grams

⇒with c(iron) = the specific heat of iron = 0.449 J/g°C

⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C

⇒with m(gold) = the mass of gold= 60.0 grams

⇒with c(gold) = the specific heat of gold = 0.128 J/g°C

⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C

60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )

26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)

26.94T2 - 6735 = -7.68T2 + 345.6

34.62T2 = 7080.6

T2 = 204.5 °C

The final temperature at the equilibrium is 204.6 °C

5 0
3 years ago
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