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2 years ago

Flammable liquids are those that have a flashpoint of:.

Chemistry

1 answer:

marin [14]2 years ago

5 0

Answer:

Flammable liquid is any liquid having a flashpoint at or below 199.4 °F (93 °C).

Explanation:

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True or False- The parts of a mixture keep their own properties when they are combined.

ICE Princess25 [194]

It is true yes :) happy to help

4 0

2 years ago

I NEED HELP PLEASE, THANKS! :)

marin [14]

Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$

6 0

3 years ago

Calculate the normality of a solution containing 147 g of h2s04 in 2L of solution

babunello [35]

I am not sure plz show me the question

5 0

3 years ago

How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?

xenn [34]

Answer:

$5.225x10^{22}atoms\ I$

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

$n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2$

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

$atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I$

Best regards!

4 0

3 years ago

For the reaction below, complete the rate expression that relates the change in concentration with respect to time to the rate o

Ann [662]

Answer: Rate in terms of disappearance of $NO$ = $-\frac{1d[NO]}{2dt}$

Rate in terms of disappearance of $Cl_2$ = $-\frac{1d[Cl_2]}{1dt}$

Rate in terms of appearance of $NOCl$ = $\frac{1d[NOCl]}{2dt}$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2NO+Cl_2\rightarrow 2NOCl$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of = $-\frac{1d[NO]}{2dt}$

Rate in terms of disappearance of = $-\frac{1d[Cl_2]}{1dt}$

Rate in terms of appearance of $NOCl$ = $+\frac{1d[NOCl]}{2dt}$

5 0

2 years ago