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irinina [24]
2 years ago
8

The uppermost layer of the atmosphere

Chemistry
2 answers:
hammer [34]2 years ago
5 0
I believe it's the exosphere
Arturiano [62]2 years ago
4 0
The thermosphere is.
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Billy the friendly Robot uses 50 N of force to lift a box 2 meters in the air. How much work did Billy perform?
nexus9112 [7]

Answer:

Work done, W = 100 J

Explanation:

We have, Billy the friendly Robot uses 50 N of force to lift a box 2 meters in the air.

It is required to find the work done by Billy.

Work done by an object is given in terms of force and displacement. The formula used to find the work done is given by :

W=Fd\\\\W=50\ N\times 2\ m\\\\W=100\ J

So, the work performed by Billy is 100 J.

4 0
3 years ago
An atom of the element iron has an atomic number of 26 and an atomic weight of 56. if it is neutral, how many protons, neutrons,
guapka [62]
Proton 26
neutron 0
you can tell how many proton by the atomic number
3 0
3 years ago
Please help, I’m a bit confused, lol
Mashutka [201]

Explanation:

option A skeleton system is correct option

hope this helps you !

4 0
2 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
What are the chemical properties of CO2<br>​
Irina-Kira [14]

Answer:

Carbon dioxide is a linear covalent molecule.

Carbon dioxide is an acidic oxide and reacts with water to give carbonic acid.

CO 2 + H2O ==> H2CO3

Carbon dioxide reacts with alkalis to give carbonates and bicarbonates.

CO 2 + NaOH ==> NaHCO3 (Sodium BiCarbonate )

NaHCO3 + NaOH ==> Na2CO3 (Sodium Carbonate) + H2O

7 0
3 years ago
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