What element is represented by the chemical symbol V

How many nucleons are in an Iodine element? Okay

Lady_Fox [76]

Answer:

there are 74

Explanation:

3 0

2 years ago

In the laboratory, a student dilutes 13.5 mL of a 11.6 M hydroiodic acid solution to a total volume of 100.0 mL. What is the con

igomit [66]

Answer: The concentration of the diluted solution is 1.566M.

Explanation:

The dilution equation is presented as this: $M_{s} V_{s} =M_{d} V_{d}$ .

·M= molarity (labeled as M)

·V= volume (labeled as L)

·s= stock solution (what you started with)

·d= diluted solution (what you have after)

Now that we know what each part of the formula symbolizes, we can plug in our data.

$11.6M*13.5mL=M_{d} *100mL$

We cannot leave it like this because the volumes must be in Liters, not milliliters. To convert this, we divide the milliliters by 1000.

$13.5mL/1000=0.0135L$ $100mL/1000=0.1L$

Now that we have the conversions, let's plug them into the equation.

$11.6M*0.0135L=M_{d} *0.1L$

The only thing that we need to do now is actually solving the answer.

$M_{d} =\frac{11.6M*0.0135L}{0.1L}$ $M_{d} =1.566M$

From the work shown above, the answer is 1.566M.

I hope this helps!! Pls mark brainliest :)

7 0

1 year ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

2 years ago

Be sure to answer all parts. Carbon dioxide (CO2) is the gas that is mainly responsible for global warming (the greenhouse effec

Elena L [17]

Answer:

1.60x10⁶ billions of g of CO₂

Explanation:

Let's calculate the production of CO₂ by a single human in a day. The molar mass of glucose is 180.156 g/mol and CO₂ is 44.01 g/mol. By the stoichiometry of the reaction:

1 mol of C₆H₁₂O₆ -------------------------- 6 moles of CO₂

Transforming for mass multiplying the number of moles by the molar mass:

180.156 g of C₆H₁₂O₆ ----------------- 264.06 g of CO₂

4.59x10² g ---------------- x

By a simple direct three rule:

180.156x = 121203.54

x = 672.77 g of CO₂ per day per human

So, in a year, 6.50 billion of human produce:

672.77 * 365 * 6.50 billion = 1.60x10⁶ billions of g of CO₂

5 0

2 years ago

At 25°C, the equilibrium constant Kc for the reaction in thesolvent CCl4 2BrCl <----> Br2 + Cl2 is 0.141. If the initial c

ivolga24 [154]

Answer: The equilibrium concentration of bromine gas is 0.00135 M

Explanation:

We are given:

Initial concentration of chlorine gas = 0.0300 M

Initial concentration of bromine monochlorine = 0.0200 M

For the given chemical equation:

$2BrCl\rightleftharpoons Br_2+Cl_2$

Initial: 0.02 0.03

At eqllm: 0.02-2x x 0.03+x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}$

We are given:

$K_c=0.141$

Putting values in above equation, we get:

$0.141=\frac{x\times (0.03+x)}{(0.02-2x)^2}\\\\x=-0.96,+0.00135$

Neglecting the value of x = -0.96 because, concentration cannot be negative

So, equilibrium concentration of bromine gas = x = 0.00135 M

Hence, the equilibrium concentration of bromine gas is 0.00135 M

8 0

2 years ago

What element is represented by the chemical symbol V​

What element is represented by the chemical symbol V