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2 years ago

If you cant see the image please let me know. last time i put an image someone said its blurry.

Chemistry

1 answer:

katovenus [111]2 years ago

8 0

Answer:

no , it's clear like a mirror

there's is nothing blurred

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URGENT <br> Best answer gets Brainliest!<br> Mg+ 2HCI ——> MgCI2+ H2

avanturin [10]

Answer:

Reaction type: Single displacement

Reactant: Magnesium

Product: Dihydrogen - H2

6 0

3 years ago

Which reactants would lead to a spontaneous reaction?

Igoryamba

Answer:

B. Ba2+ and Mn (nevermind its not B)

Explanation:

3 0

3 years ago

Using at least two properties of ionic compounds, explain why cookware is not made from ionic compounds.

muminat

Answer:

They mostly dissolve in water, and they are poor conductors of heat.

7 0

2 years ago

The gas-phase decomposition of SO2Cl2, SO2Cl2(g)→SO2(g)+Cl2(g), is first order in SO2Cl2. At 600 K the half-life for this proces

Sphinxa [80]

Answer:

a. The rate constant = 0.0000030s^-1

b. The half life = 32000secs

Explanation:Please see attachment for explanation

6 0

3 years ago

Mercury and its compounds have uses from fillings for teeth (as a mixture with silver, copper, and tin) to the production of chl

Y_Kistochka [10]

Answer:

0.6410g of HgS (mercury (II) sulfide) are formed

Explanation:

First you should write the balanced chemical equation, so we have:

$Na_{2}S+Hg(NO_{3})_{2}=HgS+_{2}NaNO_{3}$

Where:

$Na_{2}S$ is the formula for the sodium sulfide

$Hg(NO_{3})_{2}$ is the formula for the mercury (II) nitrate

$HgS$ is the formula for the mercury (II) sulfide

and

$NaNO_{3}$ is the formula for the sodium nitrate

Then you should calculate the amount of each substance in each solution, so:

- For the $Na_{2}S$ :

$Molarity=\frac{n_{Na_{2}S}}{Lofsolution}$

$n_{Na_{2}S}=0.048M*0.236L$

$n_{Na_{2}S}=0.010856$ moles of $Na_{2}S$

- For the $Hg(NO_{3})_{2}$ :

$Molarity=\frac{n_{Hg(NO_{3})_{2}}}{Lofsolution}$

$n_{Hg(NO_{3})_{2}}=0.019M*0.145L$

$n_{Hg(NO_{3})_{2}}=0.002755$ moles of $Na_{2}S$

As the quantity of $Hg(NO_{3})_{2}$ is smaller than the quantity of $Na_{2}S$ . The $Hg(NO_{3})_{2}$ is the limiting reagent and you should work with this quantity, so we have:

$0.002755molesHg(NO_{3})_{2}*\frac{1molHgS}{1molHg(NO_{3})_{2}}=0.002755$ moles of HgS

And as the molar mass of the HgS is $232.66\frac{g}{mol}$ you can calculate the mass of HgS that is produced:

$0.002755molesHgS*\frac{232.66gHgS}{1molHgS}=0.6410g$ HgS

3 0

2 years ago