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tiny-mole [99]
4 years ago
8

Which of the following would be the best object for stirring hot liquids? (1 point)

Chemistry
2 answers:
Slav-nsk [51]4 years ago
8 0
A steel spoon or copper
netineya [11]4 years ago
7 0
A steel spoon would be the best I believe.
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Find the mass in grams of 0.75 moles of magnesium (Mg).
Lady bird [3.3K]

Answer:

18.22874999999973

I recommend you to round the nearest 1 d.p

Explanation:

<em>h</em><em>a</em><em>v</em><em>e</em><em> </em><em>a</em><em> </em><em>g</em><em>r</em><em>e</em><em>a</em><em>t</em><em> </em><em>d</em><em>a</em><em>y</em><em>!</em>

3 0
3 years ago
Read 2 more answers
What would the total mass of a 4.9 mole sample of Barium (Ba) be if its Molar Mass is 137g?
Aleksandr-060686 [28]

Answer:

m = 671 grams

Explanation:

Given that,

No of moles, n = 4.9

Molar mass of Barium, M = 137 g

Mass divided by molar mass is equal to no of moles. It can be given by the formula as follows :

n=\dfrac{m}{M}\\\\m=n\times M\\\\m=4.9\times 137\\\\m=671.3\ g

or

m = 671 grams

So, the total mass of the sample of Barium is 671 grams.

8 0
3 years ago
On top of Mt. Yale in Colorado (elevation 14,202 feet), you step on a scale and weigh 183 pounds, using the same scale in New Or
babymother [125]
I believe the answer is A
7 0
3 years ago
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
If you are given 12M HCL how would you make a 6M of 100mL? SHow the math and explain the process
AlladinOne [14]

dilution

V₁M₁=V₂M₂

V₁.12 = 100.6

V₁=50 ml

Add water 50 ml to 50 ml 12 M

7 0
2 years ago
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