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3 years ago

8.5 g of rubidium are reacted completely with water.

Chemistry

1 answer:

IRINA_888 [86]3 years ago

4 0

Answer:

3.4g/dm³

Explanation:

Mass concentration = mass of solute/volume of solution

According to this question;

Mass of rubidium = 8.5g

Volume of rubidium hydroxide solution = 2.5dm³

Mass conc. = 8.5/2.5

Mass conc. 3.4g/dm³.

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A gaseous hydrogen and carbon containing compound is decomposed and formed to contain 82.66% carbon and 17.34% hydrogen by mass.

Romashka [77]

Explanation:

Mole percentage of carbon = $\frac{percentage given}{molar mass of carbon}$

= $\frac{82.66}{12}$

= 6.89

Mole percentage of hydrogen = $\frac{percentage given}{molar mass of hydrogen}$

= $\frac{17.34}{1}$

= 17.34

Now, dividing mole percentage of both the atoms by 6.89.

Then, C = 1 and H = $\frac{17.34}{6.89}$ = 2.5

Hence, empirical formula is $C_{2}H_{5}$ .

As, it is given that P = 556 mm Hg. Convert mm Hg into atm as follows.

$\frac{556 mm Hg \times atm}{760 mm Hg}$

= 0.7316 atm

Volume is given as 158 mL. So, in liter volume is $\frac{158}{1000}$ equals 0.158 L.

According to ideal gas equation, PV = nRT

$0.7316 atm \times 0.158 L = \frac{mass}{molar mass} \times 0.082 atm L/mol K \times 298 K$

$0.7316 atm \times 0.158 L = \frac{0.275 g}{molar mass} \times 0.082 atm L/mol K \times 298 K$

molar mass = 58.2 g

Hence, molecular weight of $C_{2}H_{5}$ is $12 \times 2 + 5$ = 29.

Therefore, $(C_{2}H_{5})_{n}$ = 58

29 × n = 58

n = 2

Thus, molecular formula of the compound is $C_{4}H_{10}$ .

8 0

4 years ago

Methane gas and oxygen gas react to form water vapor and carbon dioxide gas. What volume of carbon dioxide would be produced by

Bas_tet [7]

Answer:

$V_{CO_2}=9.23m^3$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

Now, as the stoichiometrical factors are in terms of mole but no information about neither the temperature nor the pressure is given, by means of the Avogadro's law, one could perform the stoichiometric calculations with the given volume as both the pressure and temperature remain the same, that is:

$V_{CO_2}=9.23m^3CH_4*\frac{1m^3CO_2}{1m^3CH_4} \\\\V_{CO_2}=9.23m^3$

Such 1:1 volume relationship equals the 1:1 molar relationship given in the chemical reaction in terms of their stoichiometric coefficients, therefore, the yielded volume of carbon dioxide is also 9.23m³

Best regards.

4 0

4 years ago

Identify the power of ten that defines each of these prefixes. Input your answers as 10 x where x is the power of ten. nano- ___

Gemiola [76]

Answer: The quantity of every prefix is written below as a power of ten.

Explanation:

In the metric system of measurement, the name of multiples and subdivision of any unit is done by combining the name of the unit with the prefixes.

For Example: deka, hecto and kilo means 10, 100 and 1000 respectively. Deci, centi and milli means one-tenth, one-hundredth, and one-thousandth respectively.

The quantity of these prefixes are written as the power of 10.

For the given prefixes:

Nano: The quantity will be $10^{-9}$