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likoan [24]
3 years ago
11

Volcanoes located at subduction zones tend to release lava high in?

Chemistry
2 answers:
Zolol [24]3 years ago
6 0
The temiture in lava is <span> 700 to 1,200 °C (1,292 to 2,192 °F).</span>
serg [7]3 years ago
3 0
Volcanoes located at subduction zones tend to release lava high in encaissantes rocks.

I hope to have helped you ! :)
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I need help ASAP!!!!!!!!!!!!!!!!!!!
liraira [26]

last one? don't take my word though

Explanation:

the suns heat is related to nuclear fusion

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2 years ago
HELP!!! An experiment was conducted to determine if the intermolecular forces in two liquids affects their boiling points. Each
ExtremeBDS [4]

Answer: it would be b time taken by the glass surface to dry

Explanation:

i had took the test and i got it right

4 0
3 years ago
X square+4x+4 factories that ​
Maslowich
Answer:
(x + 2)(x + 2)

Explanation:
You need 2 numbers that times to give 4 and add to give 4. So 2 and 2.
4 0
3 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
3 years ago
A blood pressure of 145 mm Hg is considered to be high and is generally a sign of hypertension. Determine what this pressure is
Illusion [34]

Answer:

19.3317 kilopascals

Explanation:

8 0
3 years ago
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