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2 years ago

7

It the mass of a material is 46 grams and the volume of the material is 8 cm ^3 What would the density of the material to be

1 answer:

Nat2105 [25]2 years ago

3 0

5.75 Grams per cm^3

You do mass divided by volume

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Which element has 21 neutrons and 20 electrons?

mario62 [17]

Scandium has 21 neutrons

6 0

2 years ago

Read 2 more answers

True or false. elements in the same period have the same number of valence electrons

saul85 [17]

False. elements in the same period have the same number of shells while elements in the same group have the same number of valence electrons.

4 0

3 years ago

What was the average speed of the truck for this trip?

Goshia [24]

Answer:

31.67 mph

Explanation:

To calculate the average speed of the truck, we must first obtain the total distance travelled by the truck followed by the total time taken for the truck to cover the distance travelled.

The following data were obtained from the question include:

Total distance) = 30 + 45 + 50 + 65 = 190 miles

Total time = 1 + 2 +1 +2 = 6 hours

Average speed =.?

Average speed = Total distance / Total time

Average speed = 190 /6

Average speed = 31.67 mph

Therefore, the average speed of the truck is 31.67 mph

8 0

3 years ago

What mass of copper is deposited when a current of 10.0a is passed through a solution of copper(ii) nitrate for 30.6 seconds?

asambeis [7]

Data Given:

Time = t = 30.6 s

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 63.54/2 = 31.77 g

Amount Deposited = W = ?

Solution:
According to Faraday's Law,

W = I t e / F

Putting Values,

W = (10 A × 30.6 s × 31.77 g) ÷ 96500

W = 0.100 g

Result:
0.100 g of Cu²⁺ is deposited.

3 0

3 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