A group of meteorological symbols that represents the weather at a particular observing station is a

Out of 13 students how many<br> saw themselves in the<br> horoscope?

Andru [333]

Answer: Astrology is not astronomy! ... Astronomers and other scientists know that stars many light years* away have no effect on the ... Imagine a straight line drawn from Earth through the Sun and out into space way ... The 13 constellations in the zodiac. ... on the phases of the Moon), each month got a slice of the zodiac all to itself.

Explanation: Hope this helps

5 0

3 years ago

Which of these is not an example of a molecule?

r-ruslan [8.4K]

Answer:

D

Explanation:

5 0

4 years ago

The nucleus of unstable _____ of an element will decay leading to emission of radiation.question 1 options:moleculescationsanion

Marta_Voda [28]

Isotopes of same element has different number of neutrons with different masses and having same number of protons and electrons.

Radioactive isotopes are those isotopes which are radioactive in nature. The unstable nucleus results in the radioactivity process and this process will go on until the stable isotope (element) forms.

Thus, the nucleus of unstable isotopes of an element will decay leading to emission of radiation.

6 0

3 years ago

A 0.4322 g sample of a potassium hydroxide – lithium hydroxide mixture requires 27.10 mL of 0.3565 M HCl for its titration to th

Yuki888 [10]

The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.

The mass percent of lithium hydroxide can be calculated with the following equation:

$\% = \frac{m_{LiOH}}{m_{t}} \times 100$ (1)

Where:

$m_{t} = m_{KOH} + m_{LiOH} = 0.4322 g$ (2)

We need to find the mass of LiOH.

From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.

$\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}$

$0.0271 L*0.3565 \frac{mol}{L} = \eta_{LiOH} + \eta_{KOH}$

$\eta_{LiOH} + \eta_{KOH} = 9.66 \cdot 10^{-3} \:mol$

Since mol = m/M, where M: is the molar mass and m is the mass, we have:

$\frac{m_{LiOH}}{M_{LiOH}} + \frac{m_{KOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol$ (3)

Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:

$\frac{m_{LiOH}}{M_{LiOH}} + \frac{0.4322 - m_{LiOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol$

$\frac{m_{LiOH}}{23.95 g/mol} + \frac{0.4322 g - m_{LiOH}}{56.1056 g/mol} = 9.66 \cdot 10^{-3} \:mol$

Solving for $m_{LiOH}$ , we have:

$m_{LiOH} = 0.082 g$

Hence, the percent lithium hydroxide is (eq 1):

$\% = \frac{0.082 g}{0.4322 g} \times 100 = 19.0 \%$

Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.

Learn more about mass percent here:

brainly.com/question/6992535?referrer=searchResults

brainly.com/question/5840377?referrer=searchResults

I hope it helps you!

5 0

2 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago