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2 years ago

Someone plz help me :(.

Chemistry

1 answer:

DIA [1.3K]2 years ago

8 0

Answer:

B) the kinetic energy of the particles in the spoon will decrease as the thermal energy moves from the soup to spoon.

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The reaction of an acid and a base will always produce what kind of salt?

Illusion [34]

Answer:

the answer of the question is c

7 0

2 years ago

Classify the following aqueous solutions as: strong acid, weak acid, neutral, weak base, or strong base.

kirill [66]

Vinegar pH 3.2: Weak acid

Battery acid pH 0.5: Strong acid

Shampoo pH 7.0: Neutral

Ammonia pH 11.1 Strong base

4 0

1 year ago

No links please. How many moles of methane (CH4) are in 7.31x10^25 molecules?

ad-work [718]

Answer:

molar mass of methane CH4

= C + 4 H

= 12.0 + 4 x 1.008

= 12.0 + 4.032

= 16.042g/mol

7.31 x 10^25 molecules x 1 mole CH4 = 121.43 moles

6.02 x 10^23 CH4 molecules

121.43 moles CH4 are present.

Explanation:

not to certain if this is right or not.. but hope it helps!

4 0

2 years ago

Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago

How many moles are in 2.8x10^23 atoms of Calcium?

Zielflug [23.3K]

Answer: 1 mole ➡️ 6.022×10²³ atoms of si.

X mole ➡️ 2.8×10²⁴ atoms of si.

X = 2.8×10×10²³/6.022×10²³

= 28/6.022

= 4.65 moles.

Explanation:

8 0

3 years ago