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3 years ago

3. Next to each factor isted below, increase, decrease or 'same based on how you

Chemistry

1 answer:

Nata [24]3 years ago

5 0

Answer:

Raising aw tempature- increase

sweating- decrease

shivering- increase

adding ring- same

exercising- increase

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Which of the following is true of liquids and solids?

gayaneshka [121]

The answer is D. This is because liquids take up the shape of the container they are in, so it is never definite. Where as solids stay the same shape.

3 0

4 years ago

If 18.1 mLmL of 0.800 M HClM HCl solution are needed to neutralize 5.00 mLmL of a household ammonia solution, what is the molar

AnnyKZ [126]

Answer:

Concentration of ammonia = 3M

Explanation:

The reaction for the question is a neutralization reaction between HCl and NH₃

The chemical equation for the reaction is given as

HCl + NH₃ → NH₄CL

1 : 1 1

Volume of HCl = 18.1ml = 0.0181L

Volume of NH₃ = 5ml= 0.005L

Molar concentration of HCl=0.8M

Molar Concentration of NH₃ =?

Number of moles(n)= Molar concentration × Volume

n = C × V

n of HCL= 0.8 × 0.0181

n = 0.01448moles

Since mole ration for the reaction from the balanced equation above is 1:1,

n of NH₃ = 0.01448

molar concentration (C) of NH₃ = n/v

C of NH₃ = 0.01448/0.005

C of NH₃=2.896

C OF NH₃ =3M

4 0

3 years ago

just help me with 2a) but if you have a lot of time you can just give me answer for all of it so i can check my answer , thank y

stepladder [879]

Answer:

The reaction forming aluminum bromide is Al + 3Br2 → 2AlBr3

6 0

3 years ago

natural rubidium has the average mass of 85.4678 and is composed of isotopes 85 Rb(mass = 84.9117) and 87 Rb. The ratio of atoms

mote1985 [20]

Answer:

Mass of Rb-87 is 86.913 amu.

Explanation:

Given data:

Average mass of rubidium = 85.4678 amu

Mass of Rb-85 = 84.9117

Ratio of 85Rb/87Rb in natural rubidium = 2.591

Mass of Rb = ?

Solution:

The ration of both isotope is 2.591 to 1. Which means that for 2.591 atoms of Rb-85 there is one Rb-87.

For 100% naturally occurring Rb = 2.591 + 1 = 3.591

% abundance of Rb-85 = 2.591/ 3.591 = 0.722

% abundance of Rb-87 = 1 - 0.722= 0.278

84.9117 × 0.722 + X × 0.278 = 85.4678

61.306 + X × 0.278 = 85.4678

X × 0.278 = 85.4678 - 61.306

X × 0.278 = 24.1618

X = 24.1618 / 0.278

X = 86.913 amu

8 0

3 years ago

A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a

svlad2 [7]

<u>Answer:</u> The amount of oxygen gas consumed by mouse is 0.202 grams.

<u>Explanation:</u>

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = $62.364\text{ L. Torr }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol$

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g$

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

5 0

3 years ago