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ASHA 777 [7]
3 years ago
6

Sea floor spreading occurs at a mid-ocean ridge true or false

Chemistry
1 answer:
Elenna [48]3 years ago
7 0

Answer:

False

Explanation:

sea floor spreading is not consistent at all mid ocean ridges.

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The fluorocarbon compound c2cl3f3 has a normal boiling point of 47.6°c. the specific heats of c2cl3f3(l) and c2cl3f3(g) are 0.91
myrzilka [38]
<span>To raise the liquid temperature to the point of boiling take 1231.776 joules of energy. To convert to a gas takes 5320.645 joules. To raise to 108 degrees Celsius takes 1456.848 joules. Total amount of energy needed (as heat) equals 8009.269 joules or 8.009 kj.</span>
7 0
3 years ago
Is it fair that some scientists have received more credit than others for their work on the development of the periodic table? W
Gre4nikov [31]
Yes because some work harder than others to get their credit for developing the periodic table
6 0
3 years ago
Nitrogen H2 has :
brilliants [131]

Answer:

the bond is ionic

Explanation:

6 0
2 years ago
If 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced?
IceJOKER [234]

Answer:

n_{H_2O}=1.5molH_2O

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O

Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:

n_{C_2H_6}^{available}=15g*\frac{1mol}{30g} =0.50molC_2H_6\\n_{C_2H_6}^{reacted}=60.0gO_2*\frac{1molO_2}{32gO_2}*\frac{2molC_2H_6}{7molO_2} =0.536molC_2H_6

Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:

n_{H_2O}=0.50molC_2H_6*\frac{6molH_2O}{2molC_2H_6}\\\\n_{H_2O}=1.5molH_2O

Best regards.

5 0
3 years ago
Calculate the maximum volume in ml of 0.15M HCl that each of the following antacid formulations would be expected to neutralize.
vlada-n [284]

a. 34 mL; b. 110 mL

a. A tablet containing 150 Mg(OH)₂


Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O


<em>Moles of Mg(OH)₂</em> = 150 mg Mg(OH)₂ × [1 mmol Mg(OH)₂/58.32 mg Mg(OH)₂

= 2.572 mmol Mg(OH)₂


<em>Moles of HCl</em> = 2.572 mmol Mg(OH)₂ × [2 mmol HCl/1 mmol Mg(OH)₂]

= 5.144 mmol HCl


Volume of HCl = 5.144 mmol HCl × (1 mmol HCl/0.15 mmol HCl) = 34 mL HCl


b. A tablet containing 850 mg CaCO₃


CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + H₂O


<em>Moles of CaCO₃</em> = 850 mg CaCO₃ × [1 mmol CaCO₃/100.09 mg CaCO₃

= 8.492 mmol CaCO₃


<em>Moles of HCl</em> = 8.492 mmol CaCO₃ × [2 mmol HCl/1 mmol CaCO₃]

= 16.98 mmol HCl


Volume of HCl = 16.98 mmol HCl × (1 mL HCl/0.15 mmol HCl) = 110 mL HCl


5 0
3 years ago
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