Hi . I need help on this problem please help.

What is one technology scientists use to help forecast earthquakes

schepotkina [342]

Answer:

they often use a radon detector to help forecast earthquakes

7 0

3 years ago

2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'

Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

$\displaystyle M_1V_1= M_2V_2$

Where M represents molarity and V represents volume.

Let the initial concentration and unknown volume be M₁ and V₁, respectively. Let the final concentration and required volume be M₂ and V₂, respectively. Solve for V₁:

$\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}$

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
$\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}$

Convert this value to mL:
$\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}$

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

8 0

2 years ago

Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce

sergeinik [125]

Answer:

The empirical formula is = $C_4H_8O$

The formula of Valproic acid = $C_8H_{16}O_2$

Explanation:

Mass of water obtained = 0.166 g

Molar mass of water = 18 g/mol

Moles of $H_2O$ = 0.166 g /18 g/mol = 0.00922 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.00922 = 0.01844 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.01844 x 1.008 = 0.018588 g

Mass of carbon dioxide obtained = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of $CO_2$ = 0.403 g /44.01 g/mol = 0.009157 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.009157 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.009157 x 12.0107 = 0.11 g

Given that the Valproic acid only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C - Mass of H

Mass of the sample = 0.165 g

Mass of O in sample = 0.165 - 0.11 - 0.018588 = 0.036412 g

Molar mass of O = 15.999 g/mol

Moles of O = 0.036412 / 15.999 = 0.002276 moles

Taking the simplest ratio for H, O and C as:

0.01844 : 0.002276 : 0.009157

= 8 : 1 : 4

The empirical formula is = $C_4H_8O$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 4×12 + 8×1 + 16= 72 g/mol

Molar mass = 144 g/mol

So,

Molecular mass = n × Empirical mass

144 = n × 72

⇒ n = 2

The formula of Valproic acid = $C_8H_{16}O_2$

7 0

3 years ago

A 2.83 g piece of zinc (density = 7.14 g/ml) is added to a graduated cylinder that contains 12.13 ml H2O. What will be the final

Maslowich

The final volume reading on the graduated cylinder is 12.52ml

Density = mass / volume

Volume = mass / density

Volume of zinc = 2.83/7.14

Volume of zinc = 0.3963

Initial volume = 12.13 ml

Total volume = initial volume+ volume of zinc

Total volume = 12.13 + 0.3963

Total volume = 12.5263 = 12.52 ml

To know more about volume to go the given link :

brainly.com/question/11510153

3 0

2 years ago

Need little help pls

Julli [10]

Answer:

0.354 molal

Explanation:

The molarity (M) or the concentration of a solution is defined as the number of moles of a compound in the solution per liter of solution (mol/L), whereas molality, is defined as the number of moles of a compound in the solution per kg of the compound (mol/kg).

Given that the density of the solution is 1.202 g/mL, which is equivalent to 1.202 kg/L. Since the prefix mili- denotes a factor of one thousandth ( $10^{-3}$ ) and kilo- denotes a factor of one thousand ( ${10}^3$ ),

$1.202 \ \frac{g}{mL} \ = \ 1.202 \ \frac{g}{(10^{-3}) L} \ = \ 1.202 \ \frac{{10}^3g}{L} \ = \ 1.202 \ \frac{kg}{L}$ .

To calculate the corresponding molality of the solution, the formula

$Molality \ (mol/kg) \ = \ \frac{Molarity \ (mol/L)}{Density \ (kg/L)}$ is used.

Therefore,

$Molality \ = \ \frac{0.426 \ mol/L}{1.202 \ kg/L} \ = \ 0.354 \ mol/kg \ \ (3 \ s.f.)$

4 0

3 years ago