How many moles are in 68.5 liters of oxygen gas at STP?

If I go a distance of 5 meters in 30 seconds, what is my speed

Tresset [83]

Answer:

0.166 m/s

Explanation:

speed = distance / time

= 5 / 30

= 0.166 m/s

6 0

3 years ago

What does relative abundance mean

fgiga [73]

Relative species abundance is a component of biodiversity and refers to how common or rare a species is relative to other species in a defined location or community. Relative abundance is the percent composition of an organism of a particular kind relative to the total number of organisms in the area.

I don't know if your looking for this exact definition so i'm sorry if i'm wrong

4 0

3 years ago

Today's demand curve for gasoline could shift in response to a change ina.today's price of gasoline. b.the expected future price

Nimfa-mama [501]

Answer:

d.All of the above are correct.

Explanation:

The curve of demand moves left or right continuously. Income, patterns and preferences, related products prices as well as the population size and composition are the key factors causing demand change.

3 0

3 years ago

Read 2 more answers

A coffee-cup calorimeter contains 140.0 g of water at 25.1°C . A 124.0-g block of copper metal is heated to 100.4°C by putting i

Kisachek [45]

Answer:

(a) 3347 J; (b) 3043 J; (c) 58 J/K; (d) 35.5 °C

Explanation:

(a) Heat lost by copper

The formula for the heat lost or gained by a substance is

q =mCΔT

ΔT = T₂ - T₁= 30.3 °C - 100.4 °C = -70.1 °C = -70.1 K

q = 124.0 g × 0.385 J·K⁻¹g⁻¹ × (-70.1 K) = -3347 J

The negative sign shows that heat is lost.

The copper block has lost 3347 J.

(b) Heat gained by water

ΔT = 30.3 °C - 25.1 °C = 5.2 °C = 5.2 K

q = 140.0 g × 4.18 J·K⁻¹g⁻¹ × 5.2 K = 3043 J

The water has gained 3043 J.

(c) Heat capacity of calorimeter

Heat lost by Cu = heat gained by water + heat gained by calorimeter

The temperature change for the calorimeter is the same as that for the water.

ΔT = 5.2 K

$\begin{array}{rcl}\text{3347 J} & = & \text{3043 J} + C \times \text{5.2 K}\\\text{304 J} & = & 5.2C \text{ K}\\C & = & \dfrac{\text{304 J}}{\text{5.2 K}}\\\\& = & \text{58 J/K}\\\end{array}$

The heat capacity of the calorimeter is 58 J/K.

(d) Final temperature of water

$\begin{array}{rcl}\text{Heat lost by copper } + \text{Heat gained by water}& = &0 \\\text{Heat lost by copper}& = &-\text{Heat gained by water} \\m_{\text{Cu}}C_{\text{Cu}}\Delta T_{\text{Cu}}& = & -m_{\text{w}}C_{\text{w}}\Delta T_{\text{w}}\\\end{array}\\$

$\begin{array}{rcl}\text{124.0 g} \times \text{0.385 J$\cdot$K$^{-1}$g$^{-1}$}\times \Delta T_{\text{Cu}}& = & -\text{140.0 g} \times 4.18 \text{ J$\cdot$ K$^{-1}$g$^{-1}$}\times \Delta T_{\text{w}}\\\text{47.7 J$\cdot$K$^{-1}$}\times \Delta T_{\text{Cu}}& = &-\text{585 J$\cdot$ K$^{-1}$g}\times \Delta T_{\text{w}}\\\Delta T_{\text{Cu}} & = & -12.26\Delta T_{\text{w}}\\\end{array}$

$\begin{array}{rcl}\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26(\Delta T_{\text{f}} - 30.3\, ^{\circ}\text{C})\\\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26\Delta T_{\text{f}} + 371\, ^{\circ}\text{C}\\13.26\Delta T_{\text{f}} & = & 471\, ^{\circ}\text{C}\\\Delta T_{\text{f}} & = & 35.5\, ^{\circ}\text{C}\\\end{array}$

The final temperature of the water would be 35.5 °C.

7 0

3 years ago

How many grams of solid potassium hypochlorite should be added to 2.00 L of a 0.244 M hypochlorous acid solution to prepare a bu

Olin [163]

Answer:

8.2763 g

Explanation:

Considering the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution as:

pH=pKa+log[base]/[acid]

Where Ka is the dissociation constant of the acid.

Given that the acid dissociation constant = 3.5×10⁻⁸

pKa = - log (Ka) = - log (3.5×10⁻⁸) = 7.46

Given concentration of acid = [acid] = 0.244 M

pH = 6.733

So,

6.733 = 7.46+log[base]/0.244

[Base] = 0.0457 M

Given that Volume = 2 L

So, Moles = Molarity × Volume

Moles = 0.0457 × 2 = 0.0914 moles

Molar mass of potassium hypochlorite = 90.55 g/mol

Mass = Moles × Molar mass = (0.0914 × 90.55) g = 8.2763 g

7 0

3 years ago