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3 years ago

Which phrase describes an irregular galaxy?

Chemistry

2 answers:

TiliK225 [7]3 years ago

4 0

Answer:

the answer is b

Explanation:

Stels [109]3 years ago

3 0

Answer: The answer would be B. contains many young stars.

Explanation:

There is very little new star formation in Irregular galaxies. Irregular galaxies have no particular shape and are smaller than other types of galaxies.

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Please fill this formula

Maurinko [17]

1.) Na
2.) Cl ( at the second blank)
sodium metal+hydrochloric acid

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3 years ago

Please write the process (how soap is made?.......................

Mamont248 [21]

Soap is created by mixing fats and oils with a base as opposed to detergent which is created by combining chemical compounds in a mixer.

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3 years ago

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This is the chemical formula for methyl acetate: CH32CO2. Calculate the mass percent of hydrogen in methyl acetate. Round your a

WITCHER [35]

<u>Answer:</u> The mass percent of hydrogen in methyl acetate is 8 %

<u>Explanation:</u>

The given chemical formula of methyl acetate is $CH_3COOCH_3$

To calculate the mass percentage of hydrogen in methyl acetate, we use the equation:

$\text{Mass percent of hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of methyl acetate}}\times 100$

Mass of hydrogen = (6 × 1) = 6 g

Mass of methyl acetate = [(3 × 12) + (6 × 1) + (2 × 16)] = 74 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen}=\frac{6g}{74g}\times 100=8.10\%=8\%$

Hence, the mass percent of hydrogen in methyl acetate is 8 %

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3 years ago

Hydrogen has three isotopes, 1H, 2H, and 3H. What is the difference between these three isotopes?

Firdavs [7]

C... The number of neutrons ranges from 1H with 0 neutron, 2H with 1 neutron and 3H with 2 neutrons.

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3 years ago

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20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago