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3 years ago

12. Chubchub runs a 440m lap in 1.2 minutes. How many laps will Chubchub

Chemistry

1 answer:

Svet_ta [14]3 years ago

3 0

Answer:

the number of laps in the case when he run for 50 minutes is 18,333.33

Explanation:

The computation of the number of laps in the case when he run for 50 minutes is shown below:

Given that

He runs 440m lap in 1.2 minutes

So in 50 minutes he can have laps of

= 440 × 50 ÷ 1.2

= 18,333.33 laps

hence, the number of laps in the case when he run for 50 minutes is 18,333.33

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A 150.0 mL solution of 2.888 M strontium nitrate is mixed with 200.0 mL of a 3.076 M sodium fluoride solution. Calculate the mas

Lelechka [254]

Answer:

Mass SrF2 produced = 38.63 g SrF2 produced

[Na^+]: = 1.758 M

[NO3^-]: = 1.238 M

[Sr^2+] = 0.3589 M

[F^-] = 2.36*10^-5 M

Explanation:

Step 1: Data given

Volume of 2.888M strontium nitrate = 150.0 mL = 0.150 L

Volume of 3.076 M sodium fluoride = 200.0 mL = 0.200 L

Step 2 : The balanced equation

Sr(NO3)2(aq) + 2NaF(aq) → SrF2(s) + 2NaNO3(aq) → Sr2+ + 2F- + 2

Step 3: Calculate moles strontium nitrate

Moles Sr(NO3)2 = Molarity * volume

Moles Sr(NO3)2 = 2.888 M * 0.150 L

Moles Sr(NO3)2 = 0.4332 moles

Step 4: Calculate moles NaF

Moles NaF = 3.076 M * 0.200 L

Moles NaF = 0.6152 moles

It takes 2 moles F^- to precipitate 1 mole Sr^2+, so F^- is limiting.

Step 5: Calculate limiting reactant

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

NaF is the limiting reactant. It will completely be consumed (0.6152 moles).

Sr(NO3)2 is in excess. There will react 0.6152/2 = 0.3076 moles

Moles Sr^2+ precipitated by F^- = 0.3076

There will remain 0.4332 - 0.3076 = 0.1256 moles of Sr(NO3)2

Moles Sr^2+ no precipitated (left over) = 0.1256 moles

Step 6: Calculate moles SrF2

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

For 0.6152 moles NaF we have 0.6152/2 = 0.3076 moles of SrF2

Mass SrF2 produced: 0.3076 mol * 125.6 g/mol = 38.63 g SrF2 produced

Step 7: Calculate concentration of [Na+] and [NO3-]

Since both Na^+ and NO3^- are spectator ions, and the final volume is 150 ml + 200 ml = 350 ml (0.350 L), the concentrations of Na^+ and NO3^- can be calculated as follows:

[Na^+]: (200 ml)(3.076 M) = (350 ml)(x M) and x = 1.758 M

[NO3^-]: (150 ml)(2.888 M)(2) = (350 ml)(x M) = 1.238 M

Step 8: Calculate [Sr^2+] and [F^-]

[Sr^2+] = 0.1256 moles/0.350 L = 0.3589 M

To find [F^-], one needs the Ksp for SrF2. There are several values listed in the literature. I am using a value of 2x10^-10.

SrF2(s) <==> Sr^2+(aq) + 2F^-(aq)

Ksp = [Sr^2+][F^-]²

2x10^-10 = (0.3589)(x)²

x² = 5.57*10^-10

x = [F^-] = 2.36*10^-5 M

4 0

3 years ago

How many moles of salt will you be using in this recipe?

user100 [1]

5 molessss booo booo

6 0

3 years ago

Which of the following are sources of revenue for media companies? A. Direct sales to producers B.advertising and subscriptions

Fed [463]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers

Find the quantinum numbers n,m,l,s for the last of potassium layer pleasee help explain correctly all

Fantom [35]

Answer:

Quantum numbers of the outermost electron in potassium:

$n = 4$ .
$l = 1$ .
$m_l = 0$ .
Either $m_s = 1/2$ .

Explanation:

Refer to the electron configuration of a potassium atom. The outermost electron in a ground-state potassium atom is in the $4s$ orbital (fourth $s$ orbital.)

The quantum number $n$ (the principal quantum number) specifies the main energy shell of an electron. This electron is in the fourth main energy shell (as seen in the number four in the orbital.) Hence, $n = 4$ for this electron.

The quantum number $l$ (the angular momentum quantum number) specifies the shape ( $s$ , $p$ , $d$ , etc.) of an electron. $l = 1$ for $s\!$ orbitals (such as the one that contains this electron.

Quantum numbers $n$ and $l$ specify the shape of an orbital. On the other hand, the magnetic quantum number $m_l$ specifies the orientation of these orbitals in space.

However, $s$ orbitals are spherical. Regardless of the value of $n$ , the only possible $m_l$ value for electrons in $s\!$ orbitals is $m_l = 0$ .

The spin quantum number $m_s$ distinguishes between the two electrons in an orbital. The two possible values of $m_s \!$ are $(+1/2)$ and $(-1/2)$ . Typically, the first electron in an orbital is assigned an upward ( $\uparrow$ ) spin, which corresponds to $m_s = (+1/2)$ .

5 0

3 years ago

600 s after initiation of a first order reaction 48.5% of the initial reactant concentration remains present. What is the rate c

Ludmilka [50]

Answer:

$k=1.20x10^{-3} s^{-1}$

Explanation:

For a first order reaction the rate law is:

$v=\frac{-d[A]}{[A]}=k[A]$

Integranting both sides of the equation we get:

$\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt$

where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.

From that integral we get the integrated rate law:

$ln\frac{[A]}{[A]_{0} } =-kt$

$[A]=[A]_{0}e^{-kt}$

$ln[A]=ln[A]_{0} -kt$

$k=\frac{ln[A]_{0}-ln[A]}{t}$

therefore k is

$k=\frac{ln1-ln0,485}{600}=1,20x10^{-3}$

8 0

3 years ago