The place at which two bones come together is called a

Is my answer correct?

ioda

Yes you are right .. X is undergoing a change from solid to gas at point E

7 0

3 years ago

Read 2 more answers

How many moles are 1.20 x 1025 formula units of calcium iodide?

madam [21]

Answer:

19.9 mol

Explanation:

Use Avogadro’s number to convert formula units of CaI₂ to moles of CaI₂.

1 mol CaI₂ ≡ 6.022 × 10²³ formula units CaI₂

Moles of CaI₂ = 1.20 × 10²⁵ × (1 /6.022 × 10²³)

Moles of CaI₂ = 19.9 mol

6 0

3 years ago

For the reaction: N2(g) + H2(g) → NH3(g)

gayaneshka [121]

The rate of increase of NH3 is 0.22M/s.

<h3>What is the balanced equation of the reaction?</h3>

The balanced equation of the reaction is given below:

N2(g) + 3 H2(g) → 2 NH3(g)

The rate of decrease of N2 is half the rate of increase of NH3.

Rate of decrease of N2 = -0.11 M/s

Rate of increase of NH3 = 2 × +0.11 M/s = 0.22M/s.

In conclusion, the rate of formation of products is dependent on the rate of disappearance of reactants.

Learn more about rate of reaction at: brainly.com/question/25724512

#SPJ1

3 0

2 years ago

A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.

jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

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

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $K_2CO_3$ :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of $K_2CO_3$ :

$Moles=0.200 \times {20.0\times 10^{-3}}\ moles$

Moles of $K_2CO_3$ = 0.004 moles

For $Ba(NO_3)_2$ :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of $Ba(NO_3)_2$ :

$Moles=0.400\times {30.0\times 10^{-3}}\ moles$

Moles of $Ba(NO_3)_2$ = 0.012 moles

According to the given reaction:

$K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}$

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate = 0.004 moles

Available moles of barium nitrate = 0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

(B) is correct.

4 0

3 years ago

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o

Nina [5.8K]

Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

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

The formula can be written for the calculation of moles as:

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

Thus,

$Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}$

$Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles$

$Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles$

The chemical reaction taking place:

$AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)$

According to reaction stoichiometry:

1 mole of AgNO₃ reacts with 1 mole of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

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

$Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}$

$Molarity\ of\ NaCl= 0.3374 M$

Thus, the concentration of NaCl = 0.3374 M

6 0

3 years ago