4He 2

Elements can be represented by chemical symbols. Which of the following is the correct way to write the chemical symbol for sodi

mixer [17]

Na

hope this helps :)

4 0

2 years ago

What is the volume of 18.9 g of a liquid that has a density of 0.956 g/ml

Alex_Xolod [135]

The volume is 19.76987448 by taking the known variables mass=18.9g and density=0.956g/ml
To get volume you divide the mass by the density which gives you about
19.77 ml in volume

6 0

3 years ago

The products obtained from hydroboration-oxidation of cis-2-butene are identical to the products obtained from hydroboration-oxi

Rudik [331]

Answer:

a) Attached below

b) The presence of racemic mixture found as product in both cases shows that products are identical ( i.e. they have same configuration

Explanation:

Diagrams of the products obtained from hydroboration-oxidation of cis-2-butene , hydroboration-oxidation of trans-2-butene.

attached below

The presence of racemic mixture found as product in both cases shows that products are identical ( i.e. they have same configuration )

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

2 years ago

Which combination of reactants would result in a neutralization reaction with sodium nitrate,

Debora [2.8K]

Answer:

(B) HNO3 + NaOH

Explanation:

Hello,

In this case, for the production of sodium nitrate a substance having the sodium as a cation should react with another substance having the nitrate anion. Moreover, neutralization reactions are carried out when bases react with acids, such is the case of nitric acid and sodium hydroxide:

$NaOH+HNO_3\rightarrow NaNO_3+H_2O$

Thus, the answer is (B) HNO3 + NaOH as the other options involve salts rather than acids or bases as starting reactants.

Regards.

7 0

3 years ago