Suppose that you measured out 3.50g of Na2SO4. how many moles would you have? (show work)

Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo

Romashka-Z-Leto [24]

Answer:

$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

Explanation:

Volume of a cone:

$\displaystyle V=\frac{1}{3} \pi r^2 h$

We have $\displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec}$ and we want to find $\displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ?$ when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

$\displaystyle V =\frac{1}{3} \pi (5h)^2 h$
$\displaystyle V =\frac{1}{3} \pi \ 25h^3$

Differentiate this equation with respect to time t.

$\displaystyle \frac{dV}{dt} =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}$
$\displaystyle \frac{dV}{dt} =25 \pi h^2 \ \frac{dh}{dt}$

Plug known values into the equation and solve for dh/dt.

$\displaystyle 10 = 25 \pi (2)^2 \ \frac{dh}{dt}$
$\displaystyle 10 = 100 \pi \ \frac{dh}{dt}$

Divide both sides by 100π to solve for dh/dt.

$\displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}$
$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

The height of the cone is increasing at a rate of 1/10π cm per second.

7 0

2 years ago

What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actua

notka56 [123]

Answer:

(B) 42.1%

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For $Na_2S$

Given mass = 15.5 g

Molar mass of $Na_2S$ = 78.0452 g/mol

Moles of $Na_2S$ = 15.5 g / 78.0452 g/mol = 0.1986 moles

Given: For $CuSO_4$

Given mass = 12.1 g

Molar mass of $CuSO_4$ = 159.609 g/mol

Moles of $CuSO_4$ = 12.1 g / 159.609 g/mol = 0.0758 moles

According to the given reaction:

$Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS$

1 mole of $Na_2S$ react with 1 mole of $CuSO_4$

0.1986 mole of $Na_2S$ react with 0.1986 mole of $CuSO_4$

Available moles of $CuSO_4$ = 0.0758 moles

Limiting reagent is the one which is present in small amount. Thus, $CuSO_4$ is limiting reagent. (0.0758 < 0.1986)

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

1 mole of $CuSO_4$ gives 1 mole of $CuS$

0.0758 mole of $CuSO_4$ gives 0.0758 mole of $CuS$

Molar mass of $CuS$ = 95.611 g/mol

Mass of $CuS$ = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g

Theoretical yield = 7.2473 g

Given experimental yield = 3.05 g

% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %

Option B is correct.

6 0

3 years ago

1. If you were working for a beverage company, which molarity would you recommend, and what directions for mixing would you crea

emmasim [6.3K]

Answer: I would prepare a standard solution

Explanation: A standard solution is a solution of known concentration (molarity). This is gotten from the stock using dilution principle. (C1V1=C2V2) I would prepare a 0.1M solution of the beverage into a standard 500ml flask.

7 0

2 years ago

**PLEASE SHOW YOUR WORK**Calculate how much energy would be necessary to convert 250.0 grams of ice at -30°C to steam at 145°C.

garri49 [273]

Answer:

Q= mc∆T(ice) + mLF(ice) + mc∆T(water) + mLV(water) + mc∆T(steam)

m=250 g = 0.25 kg = ¼ kg c(ice)= 2100 J/kg.K c(water)= 4200 J/kg.K LF(ice)= 333.7 kJ/kg LV(water)= 2256 kJ/kg c(steam)= 2080 J/kg.K

Explanation:

Q= ¼ × 2100 × (0°-(-30°)) + ¼ × 333700 + ¼ × 4200 × (100°-0°) + ¼ × 2256000 + ¼ × 2080 × (145°-100°)

Q= 15750 + 83425 + 105000 + 564000 + 23400

Q= 791575 J

7 0

2 years ago

Looking for 14-16, hispanic, curly hair, genuinely a nice guy

4vir4ik [10]

Answer:

discord got u tho. Go on discord.

Explanation:

5 0

3 years ago

Suppose that you measured out 3.50g of Na2SO4. how many moles would you have? (show work)​

Suppose that you measured out 3.50g of Na2SO4. how many moles would you have? (show work)