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3 years ago

Which of the dollowing isotopes have 8 neutrons in the nucleus? 0-15 0-17 C13 O C14

Chemistry

2 answers:

aleksandr82 [10.1K]3 years ago

6 0

C14. This is because there are 6 protons and the mass number is 14. 14-6 = 8 neutrons

Lorico [155]3 years ago

4 0

Answer:

C-14 is the correct answer on edge ^ they are right

Explanation:

i can cop a like or two ?

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A balloon at 32 °C is filled with 21 L of air. What would its volume be at a temperature of 52 °C, assuming pressure remains con

I am Lyosha [343]

Answer:C). 24L

Explanation: it works

4 0

4 years ago

At 25C the density of water is 0.997044 g/mL. Use this value to determine the percent error for the two density measurements

Gnom [1K]

Given that:

At 25C the density of water is 0.997044 g/mL.

From the information attached below, we have the following parameters.

The density of water calculation using a bottle.

Initial volume of Final volume of Mass of water Density (g/mL)

burette (mL) burette (mL) dispensed (g)

Sample 1 2.33 7.34 5.000 -----

Sample 2 7.34 12.37 5.025 -----

Sample 3 12.37 18.50 6.112 -----

Sample 4 18.50 24.57 6.064 -----

Sample 5 24.57 31.31 6.720 -----

The first thing we need to do is to determine the change in the volume of the burette in each sample from the above information.

The change in the volume of the burette = (final volume - the initial volume) mL

Sample 1:

= (7.34 - 2.33) mL

= 5.01 mL

Sample 2:

= (12.37 - 7.34) mL

= 5.03 mL

Sample 3:

= (18.50 - 12.37) mL

= 6.03 mL

Sample 4:

= (24.57 - 18.50) mL

= 6.07 mL

Sample 5:

= (31.31 - 24.57) mL

= 6.74 mL

The mass of the water dispersed in sample 1 is given as = 5.000 g

Using the relation for calculating the density of each, we have:

Sample 1

$\mathbf{density = \dfrac{mass}{volume}}$

$\mathbf{density = \dfrac{5.01 g}{5.000 ml}}$

density = 0.998004 g/ml

Sample 2:

$\mathbf{density = \dfrac{5.025 g}{5.03ml}}$

density = 0.999006 g/ml

Sample 3:

$\mathbf{density = \dfrac{6.112 g}{6.13ml}}$

density = 0.997064 g/ml

Sample 4:

$\mathbf{density = \dfrac{6.064 \ g}{6.07 \ ml}}$

density = 0.999012 g/ml

Sample 5:

$\mathbf{density = \dfrac{6.720 \ g}{6.74 \ ml}}$

density = 0.997033 g/ml

Thus, the average density for all the samples is:

$\mathbf{= \dfrac{( 0.998004 + 0.999006 + 0.997064 + 0.999012 + 0.997033 )}{5}}$

= 0.998024

∴

The percentage error for the two densities measurement is:

$=\dfrac{ (experimental \ value -theoretical \ value)\times 100 }{theoretical \ value}$

Given that the theoretical value = 0.997044 g/ml

Then;

$\mathbf{= \dfrac{(0.998024 - 0.997044)100}{0.997044}}$

= 0.0983%

Therefore, we can conclude that the percent error for the two density measurements is 0.0983%

Learn more about density here:

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4 0

2 years ago

This is a solution whose solvent is water.

Anastasy [175]

It would be the Aqueous solution

8 0

3 years ago

) determine the henry's law constant for ammonia in water at 25°c if an ammonia pressure of 0.022 atm produces a solution with a

Nataly [62]

Answer:

a. 59 m/atm

Explanation:

To solve this problem, we must mention Henry's law.

<em>Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.</em>

It can be expressed as: C = KP,

C is the concentration of the solution (C = 1.3 M).

P is the partial pressure of the gas above the solution (P = 0.022 atm).

K is the Henry's law constant (K = ??? M/atm),

∵ C = KP.

∴ K = C/P = (1.3 M)/(0.022 atm) = 59.0 M/atm.

3 0

3 years ago

Suppose a compound is involved in three different reactions denoted R1, R2, and R3. Tripling the concentration of this reactant

pickupchik [31]

Answer:

The order of reaction is as follows, R1 = 1; R2 = 2; R3 = 0

Explanation:

The rate of a chemical reaction is the number of moles of reactants consumed per unit time or the number of moles of products formed per unit. the rate of a chemical reaction is affected by the concentration of reactants

The relationship between the rate of a chemical reaction and the concentration of its reactants is given by the rate law or equation.

Generally, the rate equation is given as;

Rate = k[A]ᵃ[B]ᵇ..., where k = rate constant which is independent of concentration of the reactants, [A] = concentration of reactant A, a = order of reaction A, [B] = concentration of reaction B, b = order of reaction B.

For the given reactions R1, R2 and R3

For R1; rate = 3, Concentration = 3[A]

3 = k[A]3ˣ

3¹ = k[A]3ˣ

Since rate is proportional to concentration, therefore, the order of reaction, x = 1

For R2; rate = 9, Concentration = 3[A]

9 = k[A]3ˣ

3² = k[A]3ˣ

Since rate is proportional to concentration, therefore, the order of reaction, x = 2

For R1; rate = 1, Concentration = 3[A]

1 = k[A]3ˣ

3⁰ = k[A]3ˣ

Since rate is proportional to concentration, therefore, the order of reaction, x = 0

Therefore, the order of reaction is as follows, R1 = 1; R2 = 2; R3 = 0

6 0

3 years ago