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3 years ago

2. Why does the CDC recommend hand sanitizers with at least 60% alcohol?! I

Chemistry

1 answer:

choli [55]3 years ago

3 0

Answer:

Here's it:

Explanation:

Germs are everywhere! They can get onto hands and items we touch during daily activities and make us sick. Cleaning hands at key times with soap and water or hand sanitizer that contains at least 60% alcohol is one of the most important steps you can take to avoid getting sick and spreading germs to those around you.

There are important differences between washing hands with soap and water and using hand sanitizer. Soap and water work to remove all types of germs from hands, while sanitizer acts by killing certain germs on the skin. Although alcohol-based hand sanitizers can quickly reduce the number of germs in many situations, they should be used in the right situations.

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2. Use the graph below to answer the following questions.

BlackZzzverrR [31]

Answer:

The graph represents an endothermic reaction.

The products have more energy than the reactants.

80kJ

160kJ

80kJ

160kJ

6 0

2 years ago

A 35.0 ml sample of 0.225 m hbr was titrated with 42.3 ml of koh. What is the concentration of the koh?

Lemur [1.5K]

Answer:

The concentration of KOH is 0.186 M

Explanation:

First things first, we need too write out the balanced equation between HBr and KOH.

This is given as;

KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)

From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.

We use the acid base formular in calculating unknown concentrations. This is given as;

$\frac{CaVa}{CbVb} = \frac{na}{nb}$

where;

Ca = Concentration of acid

Va = Volume of acid

Cb = Concentration of base

Vb = Volume of base

na = Number of moles of acid

nb = Number of moles of base

KOH is the base and HBr is acid.

Hence;

Ca = 0.225

Va = 35

Cb = ?

Vb = 42.3

na = 1

nb = 1

Making Cb subject of formular we have;

$Cb = \frac{CaVaNb}{VbNa}$

Cb = (0.225 * 35 * 1) / (42.3 * 1)

Cb = 0.186 M

5 0

3 years ago

Is it Nephrons or Kidney or Urine or Urethra

UkoKoshka [18]

Answer:

it's urine option ( C ) .

7 0

4 years ago

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq

ad-work [718]

Answer:

-75 cm^3/min

Explanation:

Given from Boyle's law;

PV=C

From product rule;

VdP/dt + PdV/dt = dC/dt

but dC/dt = 0, V= 500 cm^3, P= 200kPa, dP= 30kPa/min

PdV/dt = dC/dt - VdP/dt

dV/dt = dC/dt - VdP/dt/ P

substituting values;

dV/dt = 0 - (500 * 30)/200

dV/dt = -75 cm^3/min

5 0

3 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago