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3 years ago

5

What is the main factor in determining the amount of thrust a rocket requires to leave the earth’s atmosphere

1 answer:

Damm [24]3 years ago

4 0

Answer:

payload weight

Explanation:

a p e x answer

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You purchase a bottle of concentrated sulfuric acid from a chemical supplier. The bottle reads Sulfuric acid (95% w/w) plastic c

My name is Ann [436]

Answer:

The molarity of the sulfuric acid is 0.018 M

Explanation:

The molarity of a solution is the number of moles of the solute (sulfuric acid in this case) in a 1-liter solution.

Every 100 g of the solution, we have 95 g sulfuric acid because its concentration is 95% w/w.

With the density, we can calculate how many liters are 100 g of solution:

density = mass / volume

1.85 g / ml = 100 g / volume

volume = 100 g / 1.85 g/ml

volume = 54.1 ml or 0.0541 l

Now, we know that we have 95 g sulfuric acid in 0.0541 l solution. In 1 l, we have then:

1 l * 95g / 0.0541 l = 1.756 g sulfuric acid.

But we want to know how many moles sulfuric acid we have per liter. Then, using the molar mass, we can calculate how many moles there are in 1.756 g sulfuric acid:

1.756 g * 1 mol / 98.08 g = 0.018 mol

The molarity is 0.018 M

4 0

3 years ago

Carbon, hydrogen, and oxygen are always found in what type of molecule?

Snowcat [4.5K]

Carbohydrates, proteins, lipids, and Nucleic Acids.

3 0

3 years ago

The compound cisplatin, pt(nh3)2cl2 , has been studied extensively as an antitumor agent.

Nataliya [291]

a. Elemental percent composition is the mass percent of each element in the compound.

The formula for mass elemental percent composition = $\frac{mass of element}{mass of compound}$ (1)

The molecular formula of cisplatin is $Pt(NH_3)_2Cl_2$ .

The atomic weight of the elements in cisplatin is:

Platinum, $Pt = 195.084 u$

Nitrogen, $N = 14.0067 u$

Hydrogen, $H = 1 u$

Chlorine, $Cl = 35.453 u$

The molar mass of $Pt(NH_3)_2Cl_2$ = $195.084+ (2\times 14.0067)+(6\times 1)+(2\times 35.453)$ = $300.00 g/mol$

The mass of each element calculated using formula (1):

- Platinum, $Pt$ %

$\frac{195.084}{300.00} \times 100 = 65.23$ %.

- Nitrogen, $N$ %

$\frac{2\times 14.0067}{300.00} \times 100 = 9.34$ %

- Hydrogen, $H$ %

$\frac{6\times 1}{300.00} \times 100 = 2.0$ %

- Chlorine, $Cl$ %

$\frac{2\times 35.453}{300.00} \times 100 = 23.63$ %

b. The given reaction of cisplatin is:

$K_2PtCl_4(aq)+2NH_3(aq)\rightarrow Pt(NH_3)_2Cl_2(s)+2KCl(aq)$

According to the balanced reaction, 1 mole of $K_2PtCl_4$ gives 1 mole of $Pt(NH_3)_2Cl_2$ .

Now, calculating the number of moles of $K_2PtCl_4$ in 100.0 g.

Number of moles = $\frac{given mass}{Molar mass}$

Molar mass of $K_2PtCl_4$ = $2\times 39.0983+195.084+4\times 35.453 = 415.093 g/mol$

Number of moles of $K_2PtCl_4$ = $\frac{100 g}{415.093 g/mol} = 0.241 mole$ .

Since, 1 mole of $K_2PtCl_4$ gives 1 mole of $Pt(NH_3)_2Cl_2$ . Therefore, mass of cisplatin is:

$0.241 mole\times 300 g/mol = 72.3 g$

For mass of $KCl$ :

Molar mass of $KCl$ = $39.0983 + 35.453 = 74.55 g/mol$

Since, 1 mole of $K_2PtCl_4$ gives 2 mole of $KCl$ . Therefore, mass of $KCl$ is:

$0.241 mole\times 74.55 g/mol\times 2 = 35.93 g$

5 0

2 years ago

One of the emission spectral lines for Be3+ has a wavelength of 253.4 nm for an electronic transition that begins in the state w

4vir4ik [10]

Answer:

4

Explanation:

Relationship between wavenumber and Rydberg constant (R) is as follows:

$wave\ number=R\times Z^2\frac{1}{n_1^2} -\frac{1}{n_1^2}$

Here, Z is atomic number.

R=109677 cm^-1

Wavenumber is related with wavelength as follows:

wavenumber = 1/wavelength

wavelength = 253.4 nm

$wavenumber=\frac{1}{253.4\times 10^{-9}} \\=39463.3\ cm^{-1}$

Z fro Be = 4

$39463.3=109677\times 4^2(\frac{1}{n_1^2} -\frac{1}{5^2})\\39463.3=109677\times 16(\frac{1}{n_1^2} -\frac{1}{5^2})\\n_1=4$

Therefore, the principal quantum number corresponding to the given emission is 4.

6 0

3 years ago

Science 10 workbook

Kruka [31]

Answer:

for more details are in the pic

4 0

3 years ago