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3 years ago

Can someone help me on 1 and 2

Chemistry

2 answers:

uranmaximum [27]3 years ago

7 0

Growth is 1. Reproduction is 2.

Because in the years the boy grew that is a characteristic of life. And the Cat giving birth is reproduction because it is when new life is made

pav-90 [236]3 years ago

6 0

I believe 1 is growth and 2 is reproduction. Hope this helps.

You might be interested in

Given the reaction to 2NaOH + H2 SO4 â Na2 SO4 + 2H2 O, what is the total number of grams of NaOH needed to react completely wit

vfiekz [6]

Answer:

4 moles, 160 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $H_2SO_4$ :-

Mass of $H_2SO_4$ = 196 g

Molar mass of $H_2SO_4$ = 98 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{196\ g}{98\ g/mol}$

$Moles\ of\ Sulfuric\ acid= 2\ mol$

According to the given reaction:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

1 mole of sulfuric acid reacts with 2 moles of NaOH

So,

2 moles of sulfuric acid reacts with 2*2 moles of NaOH

Moles of NaOH must react = 4 moles

Molar mass of NaOH = 40 g/mol

<u>Mass = Moles*molar mass = $4\times 40\ g$ = 160 g</u>

7 0

2 years ago

I what process does a solid change directly into a vapor

Marina86 [1]

Answer:

solid change directly into a vapor in the process known simply as freezing

7 0

3 years ago

Describe the shapes and relative energies of the s,p,d, and f atomic orbitals?

Mariulka [41]

Relative energies are in the order:

s < p < d < f

And the shape of these orbitals are

s - spherical

p - dumbbell

d - double dumbbell

f - double double dumbbell

6 0

3 years ago

Read 2 more answers

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indi

AlexFokin [52]

Explanation:

The given cell reaction is as follows.

$In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)$

Hence, reactions taking place at the cathode and anode are as follows.

At anode ; Oxidation-half reaction : $In(s) \rightarrow In^{3+}(aq) + 3e^{-}$ ...... (1)

At cathode; Reduction-half reaction : $Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s)$ ....... (2)

Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.

Therefore, net cell reaction is as follows.

$2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}$

$3Cd^{2+}(aq) + 6e^{-} \rightarrow 3Cd(s)$

Net reaction: $2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

Thus, we can conclude that the overall cell reaction is as follows.

$2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

4 0

3 years ago

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.

Hitman42 [59]

Explanation:

The given data is as follows.

[HCOOH] = 0.2 M, [NaOH] = 2.0 M,

V = 500 ml, [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

No. of moles of benzoic acid = Molarity × Volume

= $2 \times 0.475$

= 0.095 mol

And, moles of NaOH present in the solution will be as follows.

No. of moles of NaOH = Molarity × Volume

= $2 \times 0.025$

= 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

$C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O$

Initial: 0.095 0.05 0 0

Equlbm: (0.095 - 0.05) 0 0.05

pH = $pK_{a} + log \frac{Base}{Acid}$

= $4.2 + log \frac{0.05}{0.045}$

= 4.245

For,

$HCOOH + NaOH \rightarrow HCOONa + H_{2}O$

Initial: 0.2x 2(0.5 - x) 0

Equlbm: 0.2x - 2(0.5 - x) 0 2(0.5 - x)

As,

pH = $pK_{a} + log \frac{Base}{Acid}$

4.245 = 3.75 + $log \frac{Base}{Acid}$

$log \frac{Base}{Acid}$ = 0.5

$\frac{Base}{Acid}$ = 3.162

Now,

$\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)}$ = 3.162

x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

= 0.036 L

= 36 ml (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

8 0

3 years ago