How many milliliters are in one milligram?

What is the density of carbon dioxide gas if 0.297 g occupies of 150.0 mL?

Helga [31]

P = m/v
P = 0.297g/150.0ml
P = 1.98x10^-3 g/ml

3 0

3 years ago

The density of ethanol, a colorless liquid that is commonly known as grain alcohol, is 0.798 g/ml. calculate the mass of 16.9 ml

sesenic [268]

Hey there!:

Density = 0.798 g/mL

Volume = 16.9 mL

Therefore:

Mass = density * volume

Mass = 0.798 * 16.9

Mass = 13.4862 g

4 0

3 years ago

Để trung hoà 20ml dung dịch HCl 0.1M cần 10ml dung dịch NaOH nồng độ x mol/l . Giá trị của x là

elena55 [62]

The question is: To neutralize 20ml of 0.1M HCl solution, 10ml of NaOH solution of concentration x mol/l is required. What is the value of x?

Answer: The value of x is 0.2 M.

Explanation:

Given: $V_{1}$ = 20 mL, $M_{1}$ = 0.1 M

$V_{2}$ = 10 mL, $M_{2}$ = x

Formula used is as follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.1 M \times 20 mL = x \times 10 mL\\x = 0.2 M$

Thus, we can conclude that the value of x is 0.2 M.

6 0

3 years ago

Heavy elements such as aluminum and carbon were formed by which of these phenomena? A. Big bang B. Fusion in very large star C.

Anton [14]

The answer is D. Supernova.

4 0

3 years ago

Find the mass of 3.27 x 10^23 molecules of H2SO4. Use 3 significant digits<br> and put the units.

marta [7]

Answer:

Approximately $53.3\; \rm g$ .

Explanation:

Lookup Avogadro's Number: $N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1}$ (three significant figures.)

Lookup the relative atomic mass of $\rm H$ , $\rm S$ , and $\rm O$ on a modern periodic table:

$\rm H$ : $1.008$ .
$\rm S$ : $32.06$ .
$\rm O$ : $15.999$ .

(For example, the relative atomic mass of $\rm H$ is $1.008$ means that the mass of one mole of $\rm H\!$ atoms would be approximately $1.008\!$ grams on average.)

The question counted the number of $\rm H_2SO_4$ molecules without using any unit. Avogadro's Number $N_{\rm A}$ helps convert the unit of that count to moles.

Each mole of $\rm H_2SO_4$ molecules includes exactly $(1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23}$ of these $\rm H_2SO_4 \!$ molecules.

$3.27 \times 10^{23}$ $\rm H_2SO_4$ molecules would correspond to $\displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol$ of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of $\rm H_2SO_4$ gives the mass of each mole of $\rm H_2SO_4\!$ molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

$\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the mass of approximately $0.541389\; \rm mol$ of $\rm H_2SO_4$ :

$\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}$ .

(Rounded to three significant figures.)

6 0

3 years ago