Help please... urgently! its worth 30 points

krok68 [10]

Solutions for your questions are the following:
1. remaining amount is equal to:960 g : 100% = 30 g : xx = 30 g * 100% / 960 g
= 3.125%
= 0.03125
Now, we use this formula to calculate the number of half-lives:(1/2)ⁿ = x,
so,(1/2)ⁿ = 0.03125
to calculate n, use this equation:
n*log(1/2) = log(0.03125) n = log(0.03125)/log(1/2)
= log(0.03125)/log(0.5)
= -1.505/-0.301
n=5

Ifn = 5T = 15 min

Then
L = T/nL = 15 min/5
= 3 minutes

2. You should pick filtration. It is use to separate heterogeneous mixtures just like in the problem stated.

6 0

4 years ago

4). One mole of monoclinic sulfur at 25C was placed in a constant-pressure calorimeter whose heat capacity (C) was 1620 J/K. T

andre [41]

Answer: The enthalpy change of the reaction is -243 J/mol

Explanation:

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

$q=c\times \Delta T$

where,

c = heat capacity of calorimeter = 1620 J/K

$\Delta T$ = change in temperature = $0.150^oC=0.150K$ (Change remains same)

Putting values in above equation, we get:

$q=1620J/K\times 0.15K=243J$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

$S\text{ (monoclinic)}\rightarrow S\text{ (orthorhombic)}$

We are given:

Moles of monoclinic sulfur = 1 mole

To calculate the enthalpy change of the reaction, we use the equation:

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

where,

q = amount of heat released = -243 J

n = number of moles = 1 mole

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-243J}{1mol}=-243J/mol$

Hence, the enthalpy change of the reaction is -243 J/mol

8 0

3 years ago

Four modes of sediment transport are:

netineya [11]

Here is 5

Dissolved Load - elements dissolved in solution

Suspended Load - very fine grained sediment such as clay and silt carried in suspension. The size grains that can be carried in suspension are dependent on the current velocity

Wash Load - a subset of the suspension load, extremely small particles (clay) that will remain in suspension independent of turbulence in the river

Saltation Load - particles that are temporarily carried in suspension but move by bouncing along the bottom

Bed Load - sediment that moves by rolling or sliding along the bottom. These are generally the coarser grained sediments such as sand and gravel.

5 0

3 years ago

A cough syrup contains 5.0% ethyl alcohol, c2h5oh, by mass. if the density of the solution is 0.9928 g/ml, determine the molarit

WARRIOR [948]

To answer the question above, let us a basis of the 1000 mL or 1 L.
volume = (0.9928 g/mL)(1000mL) = 992.8 g
Then, determine the mass of the alcohol by multiplying the total mass by the decimal equivalent of 5%.
mass of alcohol = 0.05(992.8 g) = 49.64 g
Then, determine the number of moles of ethyl alcohol by dividing the mass of alcohol by the molar mass (46 g/mol).
n = 49.64 g/ (46 g/mol) = 1.08 mol
Then, divide the number of moles by the volume (our basis is 1 L)
molarity = 1.08 mol/ 1 L = 1.08 M

5 0

3 years ago

What are the major species in solution when solid ammonium bromate is dissolved in water?

DaniilM [7]

The major species in solution when solid ammonium bromate is dissolved in water is shown below

6 0

3 years ago