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2 years ago

Convert 7.68cal/sec to Kcal/min

2 answers:

8 0

First, you must convert 7.68cal/sec to cal/min. To do so, multiply 7.68x60(seconds, thus making a minute). 7.68x60=460.8

Next, we must convert 460.8 cal to kcal. To do so, divide 460.8 by 1000 (460.8/1000). The result is .4608 kcal/min

sammy [17]2 years ago

6 0

Answer:

7.68 cal/sec = 0.461 kcal/min

Explanation:

Measurement = 7.68 cal/sec

Unit conversions:

1 cal (calorie) = 10⁻³ kcal (kilocalories)

1 min = 60 sec

Step 1: convert cal to kcal

$7.68 cal/sec= \frac{7.68\ cal * 10^{-3} kcal}{1\ cal} *\frac{1}{sec} = 7.68*10^{-3} kcal/sec$

Step 2: convert sec to min

$7.68*10^{-3} kcal/sec=\frac{7.68*10^{-3} kcal}{sec} *\frac{60\ sec}{1\ min} = 0.461 kcal/min$

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jan is holding an ice cube. what causes the ice to melt? thermal energy from the ice is transferred to the air. thermal energy f

loris [4]

Answer: Ice is melting due to the transfer of thermal energy from Jan's hand to ice.

Explanation: The melting of ice is a physical change and is happening when the thermal energy from Jan's hand is transferred to ice. Due to this energy transfer, the particles of ice starts to move faster and hence, making the ice melt.

In this, the physical state of ice is changing from solid to liquid state.

$H_2O(s)\rightleftharpoons H_2O(l)$

8 0

2 years ago

Read 2 more answers

How many liters of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa)?

OLga [1]

Using ideal gas equation,

P\times V=n\times R\times T

Here,

P denotes pressure

V denotes volume

n denotes number of moles of gas

R denotes gas constant

T denotes temperature

The values at STP will be:

P=100 kPa

T=293 K

R=8.314472 L kPa K⁻¹ mol⁻¹

Number of moles of gas=3.43 mole

Putting all the values in the above equation,

$V=\frac{3.43\times 8.314\times 293}{100}$

V=83.55 L

So the volume will be 83.55 L.

83.55 L of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa).

4 0

2 years ago

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

shutvik [7]

Answer:

<u>79%</u>

Explanation:

<u>1) Balanced chemical equation:</u>

2S + 3O₂ → 2SO₃

<u>2) Mole ratio:</u>

2 mol S : 3 mol O₂ : 2 mol SO₃

<u>3) Limiting reactant:</u>

Number of moles of O₂

n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

Number of moles of S:

n = 7.0 g / 32.065 g/mol = 0.2183 mol S

Ratios:

Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

<u>4) Calcuate theoretical yield (using the limiting reactant):</u>

0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃

<u>5) Yield in grams:</u>

mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g

<u>6) </u><em><u>Percent yield:</u></em>

Percent yield, % = (actual yield / theoretical yield) × 100

% = (7.9 g / 10.0 g) × 100 = 79%

6 0

2 years ago

In a chemical reaction that takes place at a fixed pressure and volume, the enthalpy change (ΔH) is –585 kJ/mol. Will this react

Molodets [167]

by sign convention Δ $H$ is negative it means an exothermic reaction where the heat is lose so the temperature decreases.

5 0

2 years ago

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Describe how you would prepare exactly 100 mL of 0.109 M picolinate buffer, pH 5.61. Possible starting materials are pure picoli

Pepsi [2]

Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>

3 0

2 years ago