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crimeas [40]
3 years ago
7

With radiocarbon dating, scientists compare an object's carbon-14 levels with _____. the carbon-14 levels of an object from the

same time period carbon-14 levels in the atmosphere the carbon-14 levels in the rock surrounding the object the object's carbon-12 levels
Chemistry
1 answer:
trapecia [35]3 years ago
5 0

Answer:carbon-14 levels in the atmosphere

Explanation:

When carrying out radiocarbon dating, the level of carbon-14 in a sample is compared with the level of carbon 14 in the atmosphere because, objects exchange carbon-14 with the atmosphere.

Comparison of the activities of carbon-14 in the atmosphere and in the sample gives the age of the sample since the half-life of carbon-14 is a constant.

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Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be
Sphinxa [80]

Answer:

P_4_{(I)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}

Explanation:

The first step is:

P_4_{(I)}+5O_2_{(g)}\rightarrow 2P_2O_5_{(g)}

Second step is:

P_2O_5_{(g)}+3H_2O_{(l)}\rightarrow 2H_3PO_4_{(l)}

Multiplying second step by 2, and adding both the steps, we get that:

P_4_{(I)}+5O_2_{(g)}+2P_2O_5_{(g)}+6H_2O_{(l)}\rightarrow 2P_2O_5_{(g)}+4H_3PO_4_{(l)}

Cancelling common species, we get that:

P_4_{(l)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}

6 0
3 years ago
10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C
Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

Q=mc\Delte T

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c =  0.385 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c =  0.903 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c =  2.42 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c =  4.18J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c =  0.128 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0
3 years ago
Convert 0.0052 meters to millimeters
amm1812

Answer:

5.2 millimeters

Explanation:

hope this helps :)

5 0
3 years ago
Read 2 more answers
Why are reactions of ionic compounds very fast?
galina1969 [7]
Ionic compounds<span> in solution react </span>faster<span> than molecular </span>compounds<span>. This </span>is <span>because </span>Ionic compounds<span> break apart to form free </span>ions. Therefore, there are no bonds<span> to break </span>so<span> the </span><span>reaction is fast</span>
3 0
3 years ago
When 12 moles of methanol (ch3oh) and 24 moles of oxygen gas react according to the chemical equation below, what is the limitin
Nataliya [291]
The balanced reaction equation for the reaction between CH₃OH and O₂ is 
                        2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l)
Initial moles        12                24
Reacted moles   12                18
Final moles          -                   6              12               24
 
The stoichiometric ratio between CH₃OH and O₂ is 2 : 3
Hence,
   reacted moles of O₂ = reacted moles of CH₃OH x (3/2)
                                      = 12 mol x 3 / 2  
                                      = 18 mol

All of CH₃OH moles react with O₂.

Hence, the limiting agent is CH₃OH. 

Excess reagent is O₂.
Amount of moles of excess reagent left = 24 - 18 mol = 6 mol
                                            
5 0
4 years ago
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