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amm1812
3 years ago
14

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

Chemistry
1 answer:
soldi70 [24.7K]3 years ago
6 0
<span>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.</span>
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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)

2)

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:

<em> Pressure exerted by the gas is 574.85 torr</em>

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 = <em>76627.1 Pa</em>

This pressure can be stated in mmHg as

76627.1 /133.3 =<em> 574.85 mmHg </em>

and also equal to<em> 574.85 torr</em>

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]

<u>Answer:</u> The final concentration of potassium nitrate is 5.70\times 10^{-6}M

<u>Explanation:</u>

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)

  • <u>Calculating for first dilution:</u>

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

  • <u>Calculating for second dilution:</u>

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
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