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3 years ago

13

You are explaining how heat transfers. Your friend insists that when you touch something cold, you become cold

1 answer:

Dmitrij [34]3 years ago

8 0

Answer:

its C

Explanation:

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Enzymes in your body act as a catalyst. Thus the role of enzymes is to

Arturiano [62]

Answer:

o increase the rate of chemical reaction

Explanation:

Enzymes are chemical catalysts that speed up chemical reactions by lowering their activation energy. Enzymes have an active site with a unique chemical environment that fits particular chemical reactants for that enzyme, called substrates. Enzymes and substrates are thought to bind according to an induced-fit model.

6 0

3 years ago

Read 2 more answers

What discovery led J.J. Thomson to decide that atoms were not indivisible, but are actually composed of smaller parts?

Zarrin [17]

Thomson used a beam of negatively charged particles. Using a beam of particles and detecting the scattering of the particles after they hit gold foil.

7 0

4 years ago

The sulfur dioxide (so2) stack-gas concentration from fossil-fuel combustion is 12 ppmv. determine the stack-gas so2 concentrati

sergejj [24]

The concentration of $SO_{2}$ in the stack gas = 12 ppmv

That means 12 L of $SO_{2}$ is present per $10^{6} L gas$

The given temperature is 273 K (0 C) and pressure is 1 atm. At these conditions, 1 mol of gas would occupy,

$PV = nRT$

$(1 atm) (V) = (1 mol)(0.08206\frac{L.atm}{mol.K}) (273 K)$

V = 22.4 L

1 mol $SO_{2}$ occupies 22.4 L

Moles of $SO_{2}$ = $12 L *\frac{1 mol}{22.4 L} = 0.5357 mol SO_{2}$

Mass of $SO_{2}$ = $0.5357 mol *\frac{64.06 g}{1 mol} = 34.32 g SO_{2} *\frac{10^{6} microgram}{1 g}$ = $3.432 *10^{7}$ μg

Converting $10^{6} L to m^{3}$ :

$10^{6} L *\frac{1 m^{3}}{1000 L}$ = $10^{3} m^{3}$

Calculating the concentration in μg/ $m^{3}$ :

$\frac{3.432 * 10^{7} microgram}{10^{3} L} = 3.432 * 10^{4} microgram/m^{3}$

3 0

3 years ago

The temperature at which a substance in its solid state melts into the same substance in a liquid is known as

larisa86 [58]

Answer:

0 C

Explanation:

8 0

4 years ago

30 mL of 0.25 M acetic acid are titrated with 0.05 M KOH. What is the pH after addition of 75 mL KOH? Group of answer choices

hjlf

Answer:

The PH of the mixture is 4.74

Explanation:

The number of millimoles of acetic acid is calculated using the formula:

No of millimoles= Molarity * Volume( in ml)

= 0.25M * 30ml = 7.5 moles

Number of millimoles of KOH is calculated using:

Number of millimoles = Molarity * Volume ( in ml)

=0.05M * 75ml

= 3.75 moles

The PH of the solution is derived using:

pH = pKa + log [salt] / acid

= $-log [ 1.8 * 10^5 ] + log [ 3.75 mmoles/ 3.75 mmoles]$

=4.74

5 0

4 years ago