All categories

3 years ago

How would you test a colorless crystalline compound to determine if it was a hydrate?

1 answer:

6 0

A hydrate is a substance where in it contains water and other constituent elements. To know whether if that compound was a hydrate,you should record its mass, then put it in a test tube and heat it with a Bunsen burner. If the compound is a hydrate, the water in the compound will discharge in the form of water vapor. At the next 5-10 minutes, remove it in the test tube and weigh it up again. If the mass is now fewer, that means that there was water existing that has now evaporated, and the compound was a hydrate.

You might be interested in

A scientist wants to make a solution of tribasic solution phosphate, na3po4, for a laboratory experiment. How many grams of na3p

timurjin [86]

Answer:

55.75g

Explanation:

From

m/M = CV

Where

m= required mass of solute

M= molar mass of solute

C= concentration of solution

V= volume of solution=675ml

Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1

Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles

Since 1 mole of Na3PO4 contains 3 moles of Na+

It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles

mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g

3 0

3 years ago

Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 2NO2 → 2NO + O2 In a particular experiment at 300 °C, [

Setler [38]

Answer:

$rate=-1.75x10^{-5}\frac{M}{s}$

Explanation:

Hello,

In this case, for the given information, we can compute the rate of disappearance of NO₂ by using the following rate relationship:

$rate=\frac{1}{2}*\frac{C_f-C_0}{t_f-t_0}$

Whereas it is multiplied by the the inverse of the stoichiometric coefficient of NO₂ in the reaction that is 2. Moreover, the subscript <em>f</em> is referred to the final condition and the subscript <em>0</em> to the initial condition, thus, we obtain:

$rate=\frac{1}{2}*\frac{0.00650M-0.0100M}{100s-0s}\\\\rate=-1.75x10^{-5}\frac{M}{s}$

Clearly, it turns out negative since the concentration is diminishing due to its consumption.

Regards.

3 0

2 years ago

Calculate the mass of water produced when 7.26 g of butane reacts with excess oxygen

MaRussiya [10]

Answer:

11.3 g.

Explanation:

Hello there!

In this case, since the combustion of butane is:

$C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O$

Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:

$m_{H_2O}=7.26gC_4H_{10}*\frac{1molC_4H_{10}}{58.14gC_4H_{10}}*\frac{5molH_2O}{1molC_4H_{10}} *\frac{18.02gH_2O}{1molH_2O}$

Therefore, the resulting mass of water is:

$m_{H_2O}=11.3gH_2O$

Best regards!

4 0

2 years ago

What is the formula ofchromium(III) hydrogensulfate?

FromTheMoon [43]

Answer:

Cr (HSO4)3

Explanation:

its molecular weight is 343.20 g/mol

its molecular formula can also be written as CrH3O12S3

molar mass of Cr (HSO4)3 can be calculated by following method;

atomic mass of Cr = 51.9961 u

atomic mass of H = 1 u

atomic mass of S = 32.065 u

atomic mass of O = 16 u

molar mass of Cr(HSO4)3 = 51.9961+ 1.00784×3 + 32.065×3 + 15.999×12

molar mass of Cr(HSO4)3 =51.9961+3.02352+96.195+ 191.988

molar mass of Cr(HSO4)3 = 343.20 g/mol

3 0

2 years ago

How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

bezimeni [28]

Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

6 0

2 years ago