Question

a 20.4 g aluminum sphere and a 49.4 g iron sphere are both added to 75.2 mL of water contained in a graduated cylinder. what is

alexandr1967 [171]

The level in the beaker will increase because the volumes of the spheres will also be added to the volume of the water. First, we must determine the volume of each sphere. For this, we will use:

density = mass / volume
We can check the density of both aluminum and iron in literature, and given the mass, we may obtain the volume.

Aluminum:
Density = 2.70 g/ml
Mass = 20.4 g
Volume = 20.4 / 2.7 = 7.56 ml

Iron:
Density = 7.87 g/ml
Mass = 49.4 g
Volume = 49.4 / 7.87 = 6.28 ml

Now, we add these volumes to the volume of water present:
75.2 + 6.28 + 7.56 = 89.04

The new level will be 89.0 ml

3 0

3 years ago

Read 2 more answers

How are superheavy elements made?

Julli [10]

Answer: Most methods for making new elements involve a cyclotron, which speeds up atoms to high velocities before they smash into other atoms—these atoms are usually of different elements. This causes the nuclei to combine, creating new heavier elements.

Explanation: How are superheavy elements made?

3 0

2 years ago

When you combine 50.0 mL of 0.100 M AgNO3 with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, the temperature changes from

RideAnS [48]

Answer : The enthalpy of reaction $(\Delta H_{rxn})$ is, 67.716 KJ/mole

Explanation :

First we have to calculate the moles of $AgNO_3$ and $HCl$ .

$\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole$

$\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole$

Now we have to calculate the moles of AgCl formed.

The balanced chemical reaction will be,

$AgNO_3(aq)+HCl(aq)\rightarrow AgCl(s)+HNO_3(aq)$

As, 1 mole of $AgNO_3$ react with 1 mole of $HCl$ to give 1 mole of $AgCl$

So, 0.005 mole of $AgNO_3$ react with 0.005 mole of $HCl$ to give 1 mole of $AgCl$

The moles of AgCl formed = 0.005 mole

Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml

Now we have to calculate the mass of solution.

Mass of the solution = Density of the solution × Volume of the solution

Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g

Now we have to calculate the heat.

$q=m\times C\Delta T=m\times C \times (T_2-T_1)$

where,

q = heat

C = specific heat capacity = $4.18J/g^oC$

m = mass = 100 g

$T_2$ = final temperature = $24.21^oC$

$T_1$ = initial temperature = $23.40^oC$

Now put all the given values in the above expression, we get:

$q=100g\times (4.18J/g^oC)\times (24.21-23.40)^oC$

$q=338.58J$

Now we have to calculate the enthalpy of the reaction.

$\Delta H_{rxn}=\frac{q}{n}$

where,

$\Delta H_{rxn}$ = enthalpy of reaction = ?

q = heat of reaction = 338.58 J

n = moles of reaction = 0.005 mole

Now put all the given values in above expression, we get:

$\Delta H_{rxn}=\frac{338.58J}{0.005mole}=6771.6J/mole=67.716KJ/mole$

Conversion used : (1 KJ = 1000 J)

Therefore, the enthalpy of reaction $(\Delta H_{rxn})$ is, 67.716 KJ/mole

4 0

3 years ago

Commercial concentrated aqueous ammonia is 28% nh3 by mass and has a density of 0.90 g/ml. what is the concentration of ammonia

patriot [66]

Answer:- 14.9 M

Solution:- Given commercial sample of ammonia is 28% by mass. Let's say we have 100 grams of the sample. Then mass of ammonia would be 28 grams.

Density of the solution is given as 0.90 grams per mL.

From the mass and density we could calculate the volume of the solution as:

$100g(\frac{1mL}{0.90g})$

= 111 mL

Let's convert the volume from mL to L as molarity is moles of solute per liter of solution.

$111mL(\frac{1L}{1000mL})$

= 0.111 L

Now, we convert grams of ammonia to moles on dividing the grams by molar mass. Molar mass of ammonia is 17 gram per mole.

$28g(\frac{1mole}{17g})$

= 1.65 mole

To calculate the molarity we divide the moles of ammonia by the liters of solution:

$molarity=\frac{1.65mole}{0.111L}$

= 14.9 M

So, the molarity of the given commercial sample of ammonia is 14.9 M.

5 0

3 years ago

How many grams of water are produced when 4.50 L of

MA_775_DIABLO [31]

The answer for the following problem has been mentioned below.

Therefore the mass of the water is 5.802 grams.

Explanation:

Given:

volume of oxygen (V) = 4.50 L

Temperature (T) = 425 K

pressure of oxygen (P) = 2.50 atm

Gram molecular mass of oxygen (M) = 16.0 grams

To calculate:

mass of water (m)

We know;

According to the ideal gas equation;

P × V = n × R × T

As we know;

no of moles = $\frac{m}{M}$

m represents the mass of oxygen (m)

M represents the Gram molecular mass (M)

According to above mentioned equation;

P × V = n × R × T

P represents the pressure of the oxygen

V represents the volume of the oxygen

n represents the no of moles of the oxygen

R represents the universal gas constant

where,

the value of R is 0.0821 L atm/K moles

Substituting the values in the above equation;

2.50 × 4.50 = $\frac{m}{16.0}$ × 0.0821 × 425

11.25 = $\frac{m}{16.0}$ × 34.8925

180 = m × 34.8925

m = $\frac{180}{34.8925}$

m = 5.158 grams

Therefore the mass of the of oxygen is 5.158 grams

Now;

As we know;

$\frac{m_{1} }{M_{1} } = \frac{m_{2} }{M_{2} }$

where;

$m_{1}$ represents the mass of the oxygen

$M_{1}$ represents the gram molecular mass of the oxygen

$m_{2}$ represents the mass of the water

$M_{2}$ represents the gram molecular mass of water

From the above given formula,

$\frac{5.158}{16.0}$ = $\frac{m_{2} }{18}$

where;

Gram molecular weight of water = 18.0 u

$m_{2}$ = 5.802 grams

Therefore the mass of the water is 5.802 grams.

5 0

3 years ago