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Aloiza [94]
4 years ago
7

Which expression is equal to the number of milliliters(mL)in 12.5 liters(L)?

Chemistry
2 answers:
erastova [34]4 years ago
8 0
There are 1000 mL in a L.

12.5 L = 12.5(1000) mL = 12500 mL

12500 mL = 12.5 L

Hope this helps!
Cerrena [4.2K]4 years ago
4 0
12500 mL = 12.5 L would be correct
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Balance the following chemical equation: <br> ___HCO + ___O —&gt; ___H2 + ___CO3
yawa3891 [41]

Answer: 2HCO + 4O → H2 + 2CO3

Explanation: Oxomethyl + Oxygen = Dihydrogen + Carbon Trioxide

Reaction Type: SINGLE REPLACEMENT

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7 0
3 years ago
A sample of a pure compound that weighs 60.3 g contains 20.7 g Sb (antimony) and 39.6 g F (fluorine). What is the percent compos
Volgvan

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

percent composition element A=\frac{total mass of element A}{mass of compound} *100

In this case, the percent composition of fluorine is:

percent composition of fluorine=\frac{39.6 g}{60.3 g} *100

percent composition of fluorine= 65.67%

<u><em>The percent composition of fluorine is 65.67%</em></u>

4 0
4 years ago
Matter changing from a solid to a gas
DIA [1.3K]
B. sublimation is the answer.
5 0
3 years ago
The volume of a gas is 22 mL at 44°C and
sveticcg [70]

Answer:

The volume will be 18,23 ml.

Explanation:

We use the gas formula, which results from the combination of the Boyle, Charles and Gay-Lussac laws. According to which at a constant mass, temperature, pressure and volume vary, keeping constant PV / T. We convert the unit Celsius into Kelvin: 0 ° C = 273K,  44+273=317K, 17+273=290 K. The volume in L= 22/1000= 0,022L

P1xV1/T1= P2xV2/T2

V2= ((P1xV1/T1)xT2)/P2

V2=((0,905 atm x 0,022L/317K)x 290K)/0,999atm= 0,01823 L

0,01823x 1000= 18, 23 ml

5 0
3 years ago
Consider a galvanic cell based on the reaction Al^3+_(aq) + Mg_(s) rightarrow Al_(s) + Mg^2+ _(aq) The half-reactions are Al^3+
grin007 [14]

<u>Answer:</u> The standard cell potential of the cell is -0.71 V

<u>Explanation:</u>

The half reactions follows:

<u>Oxidation half reaction:</u>  Mg\rightarrow Mg^{2+}+2e^-;E^o_{Mg^{2+}/Mg}=-2.37V  ( × 3)

<u>Reduction half reaction:</u>  Al^{3+}(aq.)+3e^-\rightarrow Al(s);E^o_{Al^{3+}/Al}=-1.66V  ( × 2)

The balanced cell reaction follows:

2Al^{3+}(aq.)+3Mg(s)\rightarrow 2Al(s)+3Mg^{2+}(aq.)

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation, we get:

E^o_{cell}=-2.37-(-1.66)=-0.71V

Hence, the standard cell potential of the cell is -0.71 V

8 0
3 years ago
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