5. How many Earths could fit between the Earth and the sun?

) A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that t

sergeinik [125]

Hey there!

It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.

In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:

$m_{calcite}=15.4448g-12.4631g=2.9817g$

Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:

$V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL$

Thus, the density of the calcite sample will be:

$\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL$

This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL

Learn more:

Regards!

6 0

2 years ago

What is the molarity of 20.0 grams of NaOH dissolved in 1.50 L of solution?

Mariulka [41]

Molarity = moles/litre
mol/L=g/(g/mol) (which is g times mols per g)

will give you mols then divide that by L to get mol/L

so
M=g x mol/g x L
M=(20 x 1 x 1.50)/39.997

6 0

3 years ago

Questions 3,4,5 Brainlist

Diano4ka-milaya [45]

Answer:

3. 126.02 g of H₂S and 1109.2 g of NaI

4. 364.5g of Mg

5. 361.08 g of MgSO₄ and 108 g of H₂O

Explanation:

3. This is the reaction

2HI + Na₂S → H₂S + 2NaI

7.4 moles of HI react, so, ratio is 2:1 with H₂S and 2:2 with NaI

We would produce the same moles of NaI, 7.4 moles and the half of moles, of H₂S (7.4 /2) = 3.7 moles

Let's convert the moles to mass (mol . molar mass)

3.7 mol H₂S . 34.06 g/mol = 126.02 g

7.4 mol NaI . 149.9 g/mol = 1109.2 g

4. The balanced reaction is this:

3Mg + N₂ → Mg₃N₂

Ratio is 1:3. Therefore 1 mol of Mg₃N₂ were produced by 3 moles of Mg.

So, 5 moles of Mg₃N₂ would have been produced by 15 moles of Mg. (5 .3)

Let's convert the moles to mass (moles . molar mass)

15 mol . 24.30 g/mol = 364.5 g

5. The reaction is: H₂SO₄ + Mg(OH)₂ → MgSO₄ + 2H₂O

Ratios are 1:1 and 1:2, between sulfuric acid and the products.

If I have 3 moles of acid, I would produce 3 moles of magnessium sulfate and 6 moles of water (3 .2)

Let's convert the moles to mass

3 mol . 120.36 g/mol = 361.08 g of MgSO₄

6 mol . 18 g/mol = 108 g of H₂O

8 0

4 years ago

what electronic transition in a hydrogen atom starting from n=7 energy level will produce infrared radiation with energy of 55.1

Lostsunrise [7]

Answer:

(n = 7) ⟶ (n = 4)

Explanation:

1. Convert the energy to joules per mole of electrons.

E = 55.1 × 1000 = 55 100 J/mol

2. Convert the energy to joules per electron

E = 55 100/(6.022 × 10²³)

E = 9.150 × 10⁻²⁰ J/electron

3. Use the Rydberg equation to calculate the transition

Rydberg's original formula was in terms of wavelengths, but we can rewrite it to have the units of energy. The formula then becomes

$\Delta E = R_{\text{H}} (\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ n_{i}^{2}})$

where

$R_{\text{H}}$ = the Rydberg constant = 2.178 × 10⁻¹⁸ J

$n_{i}$ and $n_{f}$ are the initial and final energy levels.

$9.150 \times 10^{-20} = 2.178 \times 10^{-18}(\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 7^{2}})$

$\frac{9.150 \times 10^{-20} }{2.178 \times 10^{-18}} = \frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 49}$

$\frac{ 1}{ n_{f}^{2}} = \text{0.042 01} + \frac{1 }{49 }$

$\frac{ 1}{ n_{f}^{2}} = \text{0.042 01 + 0.020 41}$

$\frac{ 1}{ n_{f}^{2}} = \text{0.062 42}$

$n_{f}^{2} = \frac{1 }{ \text{0.062 42}}$

$n_{f}^{2} = 16.02$

$n_{f} = \sqrt{16.02}$

$n_{f} = 4.003 \approx 4$

4 0

3 years ago

Evaluate the following: 3.7 x 102m x 2.9 x 108 m

yulyashka [42]

Answer:

118201.68m6^2

Explanation: scientific notation

6 0

3 years ago