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3 years ago

Find the temperature from 150ml cylinder containing 0.75 moles gas at 153kpa.

1 answer:

6 0

To find the temperature in the problem, we apply the ideal gas law, PV=nRT where R=8.314 Pam3/mol K. Substituting the given, T= 153,000 Pa*1.5x10^-4 m3/ [(0.75 mol)(8.314 Pam3/mol K)]. The temperature is equal to 3.68 kelvin.

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What is the volume of 3.5 moles of oxygen gas (O2) at standard temperature and pressure (STP)?

baherus [9]

1 mole ----------- 22.4 L ( at STP )
3.5 mols --------- ?

V = 3.5 x 22.4 / 1

V = 78.4 L

hope this helps!

4 0

3 years ago

Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following

bearhunter [10]

Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the reaction:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the number of moles that we have in the 110.0 g of carbon dioxide, to this, we have to know the atomic mass of each atom:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the molar mass of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$

In other words: $1~mol~CO_2=~44~g~CO_2$ . With this in mind, we can calculate the moles:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

Now, the molar ratio between carbon dioxide and magnesium carbonate is 1:1, so:

$2.5~mol~CO_2=2.5~mol~MgCO_3$

With the molar mass of $MgCO_3$ ( $(23.3*1)+(12*1)+(16*3)=84.3~g/mol$ . With this in mind, we can calculate the grams of magnesium carbonate:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

I hope it helps!

8 0

3 years ago

Can someone help me I would really appreciate it

slega [8]

The condensed formula would be CH3-CH(CH4)-CH2-CH(CH4)-CH2-CH(CH4)-CH3. The molecular formula would be C10H25.

8 0

3 years ago

Use sedimentary in a sentence

VMariaS [17]

'The Sedimentary rock formed from years of sediments piling on top of it and being compressed.'

5 0

3 years ago

Read 2 more answers

Consider the following reaction between mercury(II) chloride and oxalate ion.

Alina [70]

Answer: The rate law of the reaction is $\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)$