FIRST ONE TO ANSWER GETS BRAINLIEST!!!!!!!!!!!HURRY!!!!!!!!!!!!!!!!!!!!!!!

If a reaction has a theoretical yield of 10.0 g and a student does the reaction and

lbvjy [14]

Answer:

78.4%

Explanation:

yield = actual/theoretical = 7.84/10 = 78.4%

6 0

3 years ago

Read 2 more answers

Calculate the number of C atoms in 0.190 mole C6H14O.<br> can anyone help me on this one?

likoan [24]

In 0.190 mole of C6H14O, there is 0.190*6 (number of C in one molecule) = 1.140 mole of C atoms. The total number of C atoms = 1.14 * 6* $10^{23}$ (atoms of C in one mole) = 6.84* $10^{23}$ atoms.

8 0

3 years ago

Commercial concentrated aqueous ammonia is 28% nh3 by mass and has a density of 0.90 g/ml. what is the concentration of ammonia

patriot [66]

Answer:- 14.9 M

Solution:- Given commercial sample of ammonia is 28% by mass. Let's say we have 100 grams of the sample. Then mass of ammonia would be 28 grams.

Density of the solution is given as 0.90 grams per mL.

From the mass and density we could calculate the volume of the solution as:

$100g(\frac{1mL}{0.90g})$

= 111 mL

Let's convert the volume from mL to L as molarity is moles of solute per liter of solution.

$111mL(\frac{1L}{1000mL})$

= 0.111 L

Now, we convert grams of ammonia to moles on dividing the grams by molar mass. Molar mass of ammonia is 17 gram per mole.

$28g(\frac{1mole}{17g})$

= 1.65 mole

To calculate the molarity we divide the moles of ammonia by the liters of solution:

$molarity=\frac{1.65mole}{0.111L}$

= 14.9 M

So, the molarity of the given commercial sample of ammonia is 14.9 M.

5 0

4 years ago

You burn a 15g jellybean to warm 50 mL of water, which increases the temperature of the water 25 °C. How many calories of heat a

Juli2301 [7.4K]

Answer:

$Q = 375\,cal$

Explanation:

The quantity of heat transfered from the jellybean to the water is:

$Q = \rho\cdot V \cdot c\cdot \Delta T$

$Q = \left(1\,\frac{g}{cm^{3}}\right)\cdot (15\,cm^{3})\cdot \left(1\,\frac{cal}{g\cdot ^{\circ} C} \right)\cdot (25\,^{\circ}C )$

$Q = 375\,cal$

7 0

3 years ago

Read 2 more answers

1. The heat of fusion for the ice-water phase transition is 335 kJ/kg at 0°C and 1 bar. The density of water is 1000 kg/m3 at th

vodomira [7]

Answer:

Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C

Explanation:

Derivation from state postulate

Using the state postulate, take the specific entropy, , for a homogeneous substance to be a function of specific volume and temperature .

ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT

During a phase change, the temperature is constant, so

ds = (partial s/partial v)(T) dv

Using the appropriate Maxwell relation gives

ds = (partial P/partial T)(v) dv

s(β) – s(aplαha) = dP/dT (v(β) – v(α))

dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv

Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.

For a closed system undergoing an internally reversible process, the first law is

du = δq – δw = Tds - Pdv

Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have

du + Pdv = dh Tds,

ds = dh/T,

Δs = Δh/T = L/T

After substitution of this result into the derivative of the pressure, one finds

dp/dT = L/TΔv

<u>This last equation is the Clapeyron equation.</u>

a)

(dP/dT) = dH/TdV => dP/dlnT = dH/dV

=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]

= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]

= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa

=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa

or

T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)

b) if the pressure in Denver is 84.6 kPa:

T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]

≅ 273.15 = 0°C T₁(freezing) essentially no change

5 0

3 years ago