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-Dominant- [34]
3 years ago
11

Write an equation for the reaction that takes place during the laboratory preparation of dry hydrogen​

Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
8 0

Answer:

Zn+2HCl→ZnCl2+H2

Explanation:

Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc.

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The statements in the table summarize what Johannes Kepler stated in the early 1600s about the motion of planets.
SIZIF [17.4K]

Answer:

it's law

Explanation:

i got it wrong because i put the wrong answer but it was law or c

i hope this help's

4 0
2 years ago
PLS HELP STOICHIOMETRY/CHEM!!!
Mariulka [41]

The molecular formula of the compound that we are required to find is the compound C4H8O8

<h3>What is empirical formula?</h3>

The empirical formula of a compound is a formula that shows the ratio of each atom present in the compound. We will start by dividing each mass with the relative atomic mass of the atom.

Carbon -  48.38 g/12   Hydrogen - 6.74 g/1    Oxygen -  53.5 g/16

Carbon - 4                    Hydrogen - 6.74         Oxygen -  8.9

Dividing through by the lowest ratio;

Carbon - 4/4            Hydrogen -  6.74/4           Oxygen 8.9/4

Carbon   1               Hydrogen    2                     Oxygen  2

The empirical formula is CH2O2.

To obtain the molecular formula; brainly.com/question/11588623

[12 + 2 + 32]n = 180

n = 180/[12 + 2 + 32]

n =4

The compound C4H8O8

Learn more about empirical formula:

7 0
2 years ago
Match each scientist to their discovery regarding the atom
neonofarm [45]

Answer:

Thomson--atoms cotain electron

Ernest Rutherford--atoms have a positive nucleus

R.A Millikan--electrons have Q=-1

Dalton--atoms are indivisible

3 0
2 years ago
Read 2 more answers
Helppppp asaapppppp plzzzzzz
Gala2k [10]

Answer:

Alright the very first thing you need to do is balance the equation:

2HCl + Na2CO3 -----> 2NaCl + CO2 + H2O

Now we need to find the limiting reactant by converting the volume to moles of both HCl and Na2CO3.

Volume x Concentration/molarity = moles

0.235L x 0.6 M = 0.141 moles / molar ratio of 2 = 0.0705 moles of HCl

0.094L x 0.75 M = 0.0705 moles /molar ratio of 1  = 0.0705 moles of Na2CO3

Since both of the moles are equal, it means the entire reaction is complete (while the identification of limiting reactant may seem like an unnecessary step, it's quite essential in stoichiometry, so keep an eye out) and there is no excess of any reactant.

Now we know that the product we want to calculate is aqueous so, following the law of conservation of mass, we should add both volumes together to calculate how much volume we could get for NaCl.

0.235 + 0.094 = 0.329L of NaCl

Now we apply the C1V1 = C2V2 equation using the concentration and volume of Na2CO3 because it's molar ratio is one to one to NaCl (You can also use HCL, but you have to divide their moles by 2 for the molar ratio)  and the volume we just calculated for NaCl.

(0.75M) x (0.094L) = C2 x (0.329L)

Rearrange equation to solve for C2:

<u>(0.75M) x (0.094L)</u>  =  C2

    (0.329L)

C2 = 0.214 M (Rounded)

<u>When the reaction is finished, the NaCl solution will have a molarity concentration of 0.214 M.</u>

<u></u>

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7 0
2 years ago
What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)
nikdorinn [45]

Answer : The volume of O_2 consumed are, 0.75\times 2\times 22.4 L.

Explanation :

The balanced chemical reaction will be:

4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)

First we have to calculate the moles of Al.

\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}

Molar mass of Al = 27 g/mole

\text{Moles of }Al=\frac{55g}{27g/mol}=2mole

Now we have to calculate the moles of O_2.

From the reaction we conclude that,

As, 4 mole of Al react with 3 moles of O_2

So, 2 mole of Al react with \frac{3}{4}\times 2=0.75\times 2 moles of O_2

Now we have to calculate the volume of O_2 consumed.

As we know that, 1 mole of substance occupies 22.4 liter volume of gas.

As, 1 mole of O_2 occupies 22.4 liter volume of O_2 gas

So, 0.75\times 2 mole of O_2 occupies 0.75\times 2\times 22.4 liter volume of O_2 gas

Therefore, the volume of O_2 consumed are, 0.75\times 2\times 22.4 L.

3 0
3 years ago
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