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3 years ago

#1: Which element has the same number of valence electrons as hydrogen (H)?

1 answer:

6 0

Valence electrons are those in the outer most orbital shell (the valence shell). Elements in the same column (group) have the same number of valence electrons. Hydrogen is in Group 1. If you were to look at a periodic table of the elements, Lithium is right under Hydrogen. You are correct, the answer is D. Lithium.

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How do the organelles within the cell allow life to occur?

spayn [35]

Answer:

There are two types of proteins: structural proteins and enzymes. Cell organelles must work together to carry out protein synthesis, utilize proteins within the cell, and transport them out of the cell.

5 0

2 years ago

Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with

stepan [7]

Answer:

1) Net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2) 0.765 M is the molarity of the carbonic acid solution.

Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

$H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq)$ ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

$KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq)$ ..[2]

In aqueous potassium carbonate , potassium ions and carbonate ion is present.:

$K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)$ ..[3]

$H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)$

From one:[1] ,[2] and [3]:

$2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)$

Cancelling common ions on both sides to get net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2CO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL$

Putting values in above equation, we get:

$M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M$

0.765 M is the molarity of the carbonic acid solution.

6 0

3 years ago

A mercury manometer is used to measure pressure in the container illustrated. Calculate the pressure exerted by the gas if atmos

Brilliant_brown [7]

Answer:

Pressure exerted by the gas is 574.85 torr

Explanation:

Atmospheric pressure = 751 torr

but 1 torr = 1 mmHg

therefore,

atmospheric pressure = 751 mmHg

1 mmHg = 133.3 Pa

therefore,

atmospheric pressure = 751 x 133.3 = 100108.3 Pa

distance labeled (tube section with mercury) = 176 mm

the pressure within the tube will be

$P_{tube}$ = ρgh

where ρ is the density of mercury = 13600 kg/m^3

h is the labeled distance = 176 mm = 0.176 m

g is acceleration due to gravity = 9.81 m/s^2

$P_{tube}$ = 13600 x 9.81 x 0.176 = 23481.216 Pa

The general equation for the pressure in the manometer will be

$P_{atm}$ = $P_{tube}$ + $P_{gas}$

where $P_{atm}$ is the atmospheric pressure

$P_{tube}$ is the pressure within the tube with mercury

$P_{gas}$ is the pressure of the gas

substituting, we have

100108.3 = 23481.216 + $P_{gas}$

$P_{gas}$ = 100108.3 - 23481.216 = 76627.1 Pa

This pressure can be stated in mmHg as

76627.1 /133.3 = 574.85 mmHg

and also equal to 574.85 torr

6 0

3 years ago

At this temperature, 0.300 mol H 2 0.300 mol H2 and 0.300 mol I 2 0.300 mol I2 were placed in a 1.00 L container to react. What

Alekssandra [29.7K]

Answer:

The concentration of HI present at equilibrium is 0.471 M.

Explanation:

5 0

3 years ago

In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

Answer: The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

Explanation:

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$ .......(1)

Calculating for first dilution:

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

Calculating for second dilution:

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

8 0

3 years ago