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NARA [144]
2 years ago
10

7.23 x 10^24 Al atoms =​

Chemistry
1 answer:
andrew11 [14]2 years ago
6 0

Answer:

12.01 moles

Explanation:

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What type of molecule is acetylacetone?
Vlada [557]

Answer:

c. ketone

Explanation:

the answer is ketone

6 0
2 years ago
A loaded coal wagon with a total mass of 20 000 kg is pushed by a force of 300 000 N. What was the wagon’s acceleration?
AlexFokin [52]

Answer:

<h2>15 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{300000}{20000}  =  \frac{30}{2}  \\

We have the final answer as

<h3>15 m/s²</h3>

Hope this helps you

5 0
3 years ago
Could somebody please give me an example of some physical properties for any substance?
KIM [24]

Answer:

A physical property is a characteristic of matter that is not associated with a change in its chemical composition. Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity.

Explanation:

6 0
2 years ago
WILL GIVE BRAINLYIST HELPPPPP PLEASE Which is the second stage of mitosis?
stich3 [128]
Metaphase is the answer
8 0
3 years ago
Read 2 more answers
A. Which reactant is the limiting reagent?
Tasya [4]

Answer:

a. Zinc is the limiting reactant.

b. m_{ZnBr_2}^{by\ Zn}=162.61gZnBr_2

c. m_{Br_2}^{leftover}=6.6g

Explanation:

Hello there!

a. In this case, when zinc metal reacts with bromine, the following chemical reaction takes place:

Zn+Br_2\rightarrow ZnBr_2

Thus, since zinc and bromine react in a 1:1 mole ratio, we can compute their reacting moles to identify the limiting reactant:

n_{Zn}=47.2g*\frac{1mol}{65.38g} =0.722molZn\\\\n_{Br_2}=122g*\frac{1mol}{159.8g} =0.763molBr_2

Thus, since zinc has the fewest moles we infer it is the limiting reactant.

b. Here, we compute the grams of zinc bromide via both reactants:

m_{ZnBr_2}^{by\ Zn}=0.722molZn*\frac{1molZnBr_3}{1molZn} *\frac{225.22gZnBr_2}{1molZnBr_2} =162.61gZnBr_2\\\\m_{ZnBr_2}^{by\ Br_2}=0.763molBr_2*\frac{1molZnBr_3}{1molBr_2} *\frac{225.22gZnBr_2}{1molZnBr_2} =171.95gZnBr_2

That is why zinc is the limiting reactant, as it yields the fewest moles of zinc bromide product.

c. Here, since just 0.722 mol of bromine would react, we compute the corresponding mass:

m_{Br_2}^{reacted}=0.722molBr_2*\frac{159.8gBr_2}{1molBr_2} =115.4gBr_2

Thus, the leftover of bromine is:

m_{Br_2}^{leftover}=122g-115.4g\\\\m_{Br_2}^{leftover}=6.6g

Best regards!

8 0
3 years ago
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