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2 years ago

2. As an airplane moves forward, the shape of its wings causes a difference in pressure

Chemistry

1 answer:

Yuki888 [10]2 years ago

7 0

Answer:

C. An upward force is created

Explanation:

As an airplane moves forward, the shape of its wing creates lower pressure above the wing than below, yielding a net upward force known as lift.

Hope this helps!

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If 35 ml of 6.0 m h2so4 was spilled, calculate the minimum mass of nahco3 that must be added to the spill to neutralize the acid

belka [17]

The balanced equation between the $H_2SO_4$ and $NaHCO_3$ is:

$H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O$

Formula of molarity is:

$Molarity = \frac{Moles of solute}{Volume of solution in Liters}$

$Molarity = 6.0 M, Volume = 35 mL = 0.035 L$

Substituting the values,

$6 = \frac{Moles of solute}{0.035}$

$Moles of solute = 0.035 L\times 6 mol/L = 0.21 mole$

So, number of moles of $H_2SO_4$ is $0.21 mole$ .

From the balanced equation it is clear that for 1 mole of $H_2SO_4$ , 2 moles of $NaHCO_3$ are required.

Hence, 0.21 mole of $H_2SO_4$ = $2\times 0.21 mole = 0.42 mole of NaHCO_3$

Molar mass of $NaHCO_3$ = $84.007 g/mol$

So, the mass of $NaHCO_3$ = $84.007 g/mol \times 0.42 mol = 35.283 g$

6 0

3 years ago

Which step of the scientific method involves obtaining resources for your question?

djyliett [7]

I would say it is constructing a hypothesis, that is because a hypothesis is basically gathering information or resources to support what your question was. so i say it is B.

3 0

3 years ago

Read 2 more answers

Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3

krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction $(\Delta S^o)$ .

$\Delta S^o=S_{product}-S_{reactant}$

$\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0{(NH_3)}$ = standard entropy of $NH_3$

$\Delta S^0{(H_2)}$ = standard entropy of $H_2$

$\Delta S^0{(N_2)}$ = standard entropy of $N_2$

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]$

$\Delta S^o=-198.3J/K$

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0

2 years ago

4. Know the location of metals, non-metals and metalloids

erastovalidia [21]

Answer:

The metals are to the left of the line (except for hydrogen, which is a nonmetal), the nonmetals are to the right of the line, and the elements immediately adjacent to the line are the metalloids.

Hope this helped

7 0

3 years ago

Characteristics of a substance that can be measured or observed by changing the identity of the material is a

algol [13]

C cause ur changing the identity of the material

4 0

2 years ago