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3 years ago

List the three great pioneers of rocket science and their contributions.

2 answers:

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Here are the three great pioneers of rocket science and some of their contributions:
1. <span>Konstantin Tsiolkovsky - a Russian teacher who realized that liquid fuel instead of solid fuel is needed for rockets to fly outside of Earth's gravity
2. Robert Goddard - an American scientist who created a new system for rockets that cooled that gases that were leaving them which improved the way rockets work significantly
3. Hermann Oberth - a German scientist who created the rocket V-2 which was used in the WWII as a missile bomb</span>

Yanka [14]3 years ago

3 0

Answer:

Here are the three great pioneers of rocket science and some of their contributions:

1. Konstantin Tsiolkovsky - a Russian teacher who realized that liquid fuel instead of solid fuel is needed for rockets to fly outside of Earth's gravity

2. Robert Goddard - an American scientist who created a new system for rockets that cooled that gases that were leaving them which improved the way rockets work significantly

3. Hermann Oberth - a German scientist who created the rocket V-2 which was used in the WWII as a missile bomb

Read more on Brainly.com - brainly.com/question/3621387#readmore

Explanation:

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Water has a slight negative charge around its oxygen atom, and a slight positive charge around its hydrogen atoms Which type of

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A student collects 350 mL of a vapor at a temperature of 67°C. The atmospheric pressure at the time of collection is 0.900 atm.

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Answer:

Explanation:

This problem is very similar to the other that you put before, so, we'll use the same principle here.

The ideal gas equation: PV = nRT

Where:

P: pressure in atm

V: volume in L

T: Temperature in K.

n: moles

R: Gas constant (In this case, we'll use 0.082 L atm/K)

to get the molar mass of the gas, we need to know the moles, and with the mass, we can know the molar mass. However we can put the ideal gas expression with the molar mass in this way:

we know that n is mole so:

n = g/MM

If we put this in the idea gases expression we have:

PV = gRT/MM

Solving for MM we have:

MM = gRT/PV

Now, let's convert the temperature and volume to K and L respectively:

T = 67 + 273 = 340 K

V = 350 / 1000 = 0.35 L

Now all we have to do is put all the data into the expression:

MM = 0.79 * 0.082 * 340 / 0.9 * 0.35

MM = 22.0252 / 0.315 = 69.92 g/mol rounded 70 g/mol

Now, the closest answer of your options would be 72 g/mol. This could be easily explained because we do not use all the significant figures of all numbers, including the gas constant of R. However, all the work, calculations and procedure is correct and fine, and we only have a minimum range of 2 units.

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1.00 M CaCl2 Density = 1.07 g/mL

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Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

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