Drag the tiles to the correct boxes to complete the pairs.

Match the descriptions with the types of blas.

selection bias

expectation bias

confirmation bias

contextual bias

paying more attention to evidence that

confirms one's hypothesis and ignoring

evidence that may discount it

to be swayed from one's conclusion by

additional information

coming to a conclusion before all the

evidence has been processed and therefore

unconsciously disregarding evidence to the

contrary

Answer: Mg oxidised to Mg++ and Ag+ reduced to Ag

Explanation:No ionic equation given!!

assume equation is Mg + 2Ag+ + 2NO3- —> Mg++ + 2Ag + 2NO3-

redox reaction: Ag gains an electron = reduced

Mg loseselectrons = oxidised

Answer:

0.229 cm³.

Explanation:

The following data were obtained from the question:

Volume (in in³) = 0.014 in³

Volume (in cm³) =?

1 in = 2.54 cm

Next, we shall determine a conversion scale to convert from in³ to cm³. This can be obtained as follow:

1 in = 2.54 cm

Therefore,

1 in³ = 2.54³ cm³

1 in³ = 16.387 cm³

Finally, we shall convert 0.014 in³ to cm³. This can be obtained as follow:

1 in³ = 16.387 cm³

Therefore,

0.014 in³ = 0.014 in³ × 16.387 cm³ / 1 in³

0.014 in³ = 0.229 cm³

Thus, 0.014 in³ is equivalent to 0.229 cm³.

Answer:

2.9moles of hydrogen gas

Explanation:

convert liters to dm³

since 1liter= 1dm³

thus, 65.0liters = 65.0dm³

number of moles = volume given/22.4dm³

= 65.0/22.4

=2.9moles

Answer:

movement of charged particles through the wire .

Explanation:

When electricity is passed through the wire of electromagnet , moving electrons of the wire produces magnetic field . This magnetic field in increased due to high permeability of soft iron of the electromagnet . It is this magnetic field which creates magnetic force .

Answer:

A. This allows the plants to conserve water and not bloom during the heat of the day

Explanation:

Most desert plants only bloom at night because they take advantage of animals like moths and insects that fly at night for pollination and reproduction.

Because these plants are in the desert and do not get enough water except from short occasional rainfalls, they conserve water and not bloom during the heat of the day. They bloom at night when the temperature is low and this enhances their water conservation and survival.

Answer:

d. 9.95 × 10⁻³ mol

e. 249 g

f. 2.55 × 10²⁴ atoms

g. 4.32 × 10²² atoms

Explanation:

d. the number of moles of cobalt represented by 5.99 x 10²¹ cobalt atoms

We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.

5.99 x 10²¹ atoms × 1 mol/6.02 × 10²³ atoms = 9.95 × 10⁻³ mol

e. the mass of 4.23 mol of cobalt

The molar mass of cobalt is 58.93 g/mol.

4.23 mol × 58.93 g/mol = 249 g

f. the number of cobalt atoms in 4.23 mol of cobalt

We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.

4.23 mol × 6.02 × 10²³ atoms/1 mol = 2.55 × 10²⁴ atoms

g. the number of cobalt atoms in 4.23 g of cobalt

First, we will calculate the moles of cobalt using the molar mass of cobalt.

4.23 g × 1 mol/58.93 g = 0.0718 mol

Then, we will calculate the number of cobalt atoms using Avogadro's number.

0.0718 mol × 6.02 × 10²³ atoms/1 mol = 4.32 × 10²² atoms

Answer:

5.78 × 10¹⁴ Hz

Explanation:

Step 1: Given and required data

Step 2: Convert "λ" to meters

We will use the conversion factor 1 m = 10⁹ nm.

519. nm × 1 m/10⁹ nm = 5.19 × 10⁻⁷ m

Step 3: Calculate the frecquency of this light

We will use the following expression.

c = λ × ν

ν = c/λ

ν = (3.00 × 10⁸ m/s)/5.19 × 10⁻⁷ m

ν = 5.78 × 10¹⁴ s⁻¹ = 5.78 × 10¹⁴ Hz

Answer: product

Explanation:

Each substance written to the right of the arrow in a chemical equation is referred to as a product.

When writing a chemical equation, the substance that's written to the left of arrow in the equation is the reactants.

On the other hand which is the right side is the product.

Answer:

In advance

middle

lower

Explanation:

These are the safety precautions needed when carrying out duties in the fume hood.

When planning and preparing to work in a fume hood (a locally designed area to reduce exposure to hazardous fumes). It is advisable to make all equipment readily available at your disposal in advance to reduce and minimize the raising and lowering of the hood sash at intervals.

It is also pertinent to understand that working in the middle of the work surface helps to promote the movement of air and keeps the area neat and tidy.

However, if any case where there is a need to get a new tool or equipment during the process of working in a fume hood, it is advisable to lower the sash at that point in time.

Answer:

Solution A: 0.00400M

Solution B: 0.00400M

Solution C: 4.00x10⁻⁵M

Explanation:

Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:

250mL / 10mL = 25 times.

That means molar concentration of sln A is:

0.100M / 25 = 0.00400M

Solution B is obtained diluting 25mL to 100mL:

100mL / 25mL = 4 times

0.00400M / 4 times = 0.00100M

And solution C is obtained diluting the solution C from 20mL to 500mL:

500mL / 20mL = 25 times

Solution C:

0.00100M / 25 times = 4.00x10⁻⁵M

The formula for serial dilution can be used to obtain the molarity of solution A, B , C.

M1V1 = M2V2

M2 = 0.100 M ×  10 mL/250-mL

M2 = 0.004 M

M1V1 = M2V2

M2 = 0.004 M × 25 mL/100-mL

M2 = 0.001 M

M1V1 = M2V2

M2 = 0.001 M × 20 mL/500-mL

M2 = 0.00004 M

Learn more about serial dilution: https://brainly.com/question/2167827

Answer:

Explanation:

Nitric oxide is the gas NO, it reacts with oxygen as shown below;

2NO(g) + O2(g) -----> 2NO2(g)

Now the gas formed is the gas NO2 which is known to be reddish brown in colour.

A diagrammatic representation of this reaction is shown in the image attached to this answer.

Image credit: Chemlibretext

Answer:

The cycling of matter is important to many Earth processes and to the survival of organisms the existing matter must cycle continuously for this planet to support life Water, carbon, nitrogen, phosphorus, and even rocks move through cycles If these materials did not cycle, Earth could not support life.

Explanation:

Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.

Rocks, as well as water, carbon, nitrogen, and phosphorus, go through cycles. The planet Earth could not support life if these materials did not cycle.

Subsystems exist within the Earth system. These subsystems include the exosphere, atmosphere, hydrosphere, lithosphere and geosphere, also referred to as the lithosphere, and the living environment (biosphere).

These systems are powered by energy that comes from both the Sun and the interior of the Earth. Through processes known as biogeochemical cycles, nutrients and elements also move through these systems along with energy.

Therefore,  Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.

To learn more about Earth, refer to the link:

https://brainly.com/question/1204146

#SPJ6

Answer:

0.57 water

Explanation:

To solve this problem, we need to write the reaction expression first.

The reactants are oxygen gas and hydrogen gas.

They react to give a product of water

       2H₂    +    O₂   →   2 H₂O  

Given that;

Number of moles of hydrogen gas = 0.57moles

From the balanced reaction expression;

       2 moles of hydrogen gas produces 2 moles of water

   So;

    0.57mole of hydrogen gas will also produce 0.57 water

Answer:

Other species will slowly die, because the species that eat snails and crayfish will not have anything to eat, then the thing that eats them will not have anything. This process will go on and on. So long story short populations will decrease, then off.

Answer:

th relative ability of a solute to devolve into a solvent

The percent composition of the compound.

A O-18.18%, N-21.21%, H-60.60%

Given

6.00 grams of oxygen,

7.00 grams of nitrogen,

20.00 grams of hydrogen.

Required

The percent composition

Solution

Total mass :

= mass of O + mass of N + mass of H

= 6 + 7 + 20

= 33 g

% O = 6/33 x 100%= 18.18%

% N = 7/33 x 100%=21.21%

% H = 20/33 x 100% = 60.6 %

Answer:

Average atomic mass  = 3.9 amu

Explanation:

Given data:

Percent abundance of He-2 = 0.420%

Percent abundance of He-3 = 2.75%

Percent abundance of He-4 = 96.83%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass)  / 100

Average atomic mass = (0.420×2)+(2.75×3) +(96.83×4)/100

Average atomic mass =  0.84 + 8.25 +387.32 / 100

Average atomic mass = 396.41 / 100

Average atomic mass  = 3.9 amu.

Answer:

In biology class we had to look at the parts of a plant in a microscope

Answer:

Observational evidence is essential for investigating the way disease affects populations, the patterns and distribution of risk within them, and the emergence of trends in health and disease over time.

Answer:

It tests a prediction It supports the results. It asks a testable question It predicts what will happen

Explanation:

Answer:

The answer is "A"

Explanation:

Please find the complete question in the attachment file.

[tex]\to R(s)= \frac{As}{K+s}[/tex]

when the s in the approach, that is infinity R(s) tends

[tex]\to \frac{A}{\frac{K}{s}+1} \\\\ \to\frac{A}{0+1} \\\\ \to\frac{A}{1} \\\\ \to A[/tex]

Answer:

hello your question is incomplete attached below is the complete question

answer :

we use Isomerization because conjugated allylic carbocation  is more stable when compared to a Non-conjugated Allylic carbocation

Explanation:

Reason for the mechanism

we use Isomerization because conjugated allylic carbocation  is more stable when compared to a Non-conjugated Allylic carbocation

attached below is the detailed mechanism

Answer:

I don't get it is it even a question?

Answer:

[N₂] = 1.1M

Explanation:

Based on the chemical reaction:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Equilibrium constant, K, is defined as:

K = 5.93 = [NO]² / [N₂] [O₂]

Where [] are equilibrium concentrations of each specie

As initial concentrations are:

N₂ = 0.40mol / 0.500L = 0.8M

NO = 1mol / 0.500L = 2M

The equilbrium concentrations are:

[NO] = 2M - 2X

[N₂] = 0.8M +X

[O₂] = X

Replacing:

5.93 = [2 - 2X]² / [0.8+X] [X]

5.93 = 4 - 8X + 4X² / 0.8X + X²

4.744X + 5.93X² = 4 - 8X + 4X²

1.93X² + 12.744X - 4 = 0

Solving for X:

X = -6.9M → False solution. There are no negative concentrations

X = 0.3M. Real solution.

[N₂] in equilibrium is:

[N₂] = 0.8M +0.3M

[N₂] = 1.1M

Answer:

V₂ = 104.76 mL

Explanation:

Given data:

Initial volume = 100.0 mL

Initial temperature = 21°C (21 + 273.15 K = 294.15 K)

Final temperature = 35°C (35 + 273.15 K = 308.15 k)

Final volume = ?

Solution:

Charles Law:

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ =100.0 mL × 308.15 K / 294.15 K

V₂ = 30815 mL.K /294.15 K

V₂ = 104.76 mL

Answer:

% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

% Free space in ice  = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

Explanation:

As given ,

Density for ice at 0⁰C = 0.917 g/ml

Density for water at 0⁰C = 0.999 g/ml

Radii of H atoms = 37 pm

Radii of O atoms = 66 pm

Now,

Consider 1 ml of water = 1 cm²

As , we know that mass of water in 1 cm² = 0.999 g

Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]

Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²

Now,

Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 5.48×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

Now,

Consider 1 ml of ice  = 1 cm²

S.I unit of ice = 1×[tex]10^{-6}[/tex] m²

As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g

Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]

Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012

Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]

Now,

Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 1.17×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

Answer:

H-H equation is written as follows:

pH=pK + log

{HCO3-}(base)

{H2CO3}(acid)

Answer: 1.2 x 10^-17 microounces

Explanation:

Ounce = 28.5G microounce = 28.5*10^-6g

3.4*10^-25 kg = 3.4*10^-22 g = (3.4/2.85)*10^(-22+5) = 1.2*10-17