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2 years ago

15

During a workout, you set the treadmill to a pace of 55.0m/min. How many minutes will you walk if you cover a distance of 7500ft

?

1 answer:

Licemer1 [7]2 years ago

8 0

55 m=180.446 ft.
7500/180.446= approx. 41.56 minutes

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What volume of 0.4567 m h2so4 is required to neutralize 30.00 ml of 0.3210 m naoh?

sertanlavr [38]

Answer is: volume of H₂SO₄ is 42.1 mL.<span>
Chemical reaction: H</span>₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.<span>
c(H</span>₂SO₄) = 0,4567 M = 0,4567 mol/L.<span>
V(NaOH) = 30 mL </span>÷ 1000 mL/L <span>= 0,03 L.
c(NaOH) = 0,321 M = 0,321 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,321 mol/L · 0,030 L.
n(NaOH) = 0,00963 mol.
From chemical reaction: n(H</span>₂SO₄) : n(NaOH) = 1 : 2.<span>
n(H</span>₂SO₄) = 0,01926 mol.<span>
V(H</span>₂SO₄) = n(H₂SO₄) ÷ c(H₂SO₄).<span>
V(H</span>₂SO₄) = 0,01926 mol ÷ 0,4567 mol/L.<span>
V(H</span>₂SO₄<span>) = 0,0421 L = 42,1 mL.</span>

3 0

2 years ago

Refer to the following table for information about gold and mercury. Then choose which statements are true about equal sample si

babymother [125]

Below are the choices:

A The mercury will change temperature at a much faster rate under the same heating conditions.
<span>B The two metal samples will change temperature at about the same rate. </span>
<span>C The gold would float if placed in the mercury. </span>
<span>D The gold would sink to the bottom if placed in the mercury.
</span>
<span>a = false, it will take 0.031 cal to raise 1g Au 1degree while it will take 0.033 cal to raise 1g Hg 1 degree so, although Au will heat up faster, it will not be discernably faster so...
b = true
c = false, Au density > Hg
d = true</span>

7 0

3 years ago

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

2 years ago

What is the thermal energy needed to vaporize 18.02g of water at 100°C? The Heat of Vaporization for water is 40.650 kJ/mol.

podryga [215]

Answer:

40.7 kJ

Explanation:

Applying,

q = c'n.................. Equation 1

Where q = Thermal Heat, c' = Heat of vaporization of water, n = number of mole of water.

But,

n = mass(m)/Molar mass(m')

n = m/m'............... Equation 2

Substitute equation 2 into equation 1

q = c'(m/m')............. Equation 3

Given: c' = 40.650 KJ/mol, m = 18.02 g

Constant: m' = 18 g/mol

Substitute into equation 3

q = 40.650(18.02/18)

q = 40.695 kJ

q ≈ 40.7 kJ

5 0

2 years ago

What refers to the effect of gravity

Grace [21]

Answer: D

Explanation:

Weight by definition is mass multiplied by the acceleration of gravity. So, weight is the force of gravity

4 0

2 years ago