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2 years ago

Write a hypothesis for this: Hummingbirds are attracted to the color red

Chemistry

1 answer:

Margaret [11]2 years ago

8 0

Answer: I believe/don't believe hummingbirds are attracted to the color red because ... (Enter your reasons)

Explanation:

Hypothesis are always different, up to you to choose believe or not believe depending on your position.

:) Hope this helped.

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Explain the observed pattern of how the sizes and charges of atoms change with the addition and subtraction of electrons. ( NEED

siniylev [52]

Atoms are the smallest particles of matter.
b. The mass of an iron atom is different from the mass of a copper atom.
c. Every atom of silver is identical to every other atom of silver.
d. A compound is composed of atoms of two or more different elements.

3 0

2 years ago

In Males which of the following organ performs a dual function of ejaculation of sperm and voiding of urine

Anna11 [10]

I believe it is the urethra

3 0

2 years ago

Read 2 more answers

A 14.3-cm3 sample of tin has a mass of 0.104 kg.

VladimirAG [237]

This is a true statement if it is density you are looking for... Density problem.....

Density is the ratio of the mass of an object to its volume.
D = m / V
D = 104g / 14.3 cm³ = 7.27 g/cm³ .............. to three significant digits

The conventions for the units of density is that grams per cubic centimeter (g/cm³) are usually used for solids, but will work for anything. Grams per milliliter (g/mL) are usually used for liquids and grams per liter (g/L) are for gases. Therefore, by convention, the units for tin (a solid) should be in grams per cubic centimeter.

Since 1 mL is equivalent to 1 cm³, then the density could be expressed as 7.27 g/mL.

The accepted value for the density of tin is 7.31 g/cm³

7 0

2 years ago

When 70. milliliter of 3.0-molar Na2CO3 is added to 30. milliliters of 1.0-molar NaHCO3 the resulting concentration of Na+ is 2

tiny-mole [99]

Answer : The resulting concentration of $Na^+$ ion is, 4.5 M

Explanation : Given,

Concentration of $Na_2CO_3$ = $M_1$ = 3.0 M = 3.0 mol/L

Volume of $Na_2CO_3$ = $V_1$ = 70 mL = 0.07 L

Concentration of $NaHCO_3$ = $M_2$ = 1.0 M = 1.0 mol/L

Volume of $NaHCO_3$ = $V_2$ = 30 mL = 0.03 L

First we have to calculate the moles of $Na_2CO_3$ and $NaHCO_3$

$\text{Moles of }Na_2CO_3=\text{Concentration of }Na_2CO_3\times \text{Volume of }Na_2CO_3=3.0mol/L\times 0.07L=0.21mol$

and,

$\text{Moles of }NaHCO_3=\text{Concentration of }NaHCO_3\times \text{Volume of }NaHCO_3=1.0mol/L\times 0.03L=0.03mol$

Now we have to calculate the moles of $Na^+$ ions.

As, 1 mole of $Na_2CO_3$ will give 2 moles of $Na^+$ ions

So, 0.21 moles of $Na_2CO_3$ will give $2\times 0.21=0.42$ moles of $Na^+$ ions

and,

As, 1 mole of $NaHCO_3$ will give 1 mole of $Na^+$ ions

So, 0.03 moles of $NaHCO_3$ will give 0.03 moles of $Na^+$ ions

So,

Total number of moles of $Na^+$ ions = 0.42 + 0.03 =0.45 mole

Total volume of both solution = 70 mL + 30 mL = 100 mL = 0.1 L

Now we have to calculate the concentration of $Na^+$ ions.

$\text{Concentration of }Na^+=\frac{\text{Moles of }Na^+}{\text{Volume of solution}}=\frac{0.45mol}{0.1L}=4.5mol/L=4.5M$

Therefore, the resulting concentration of $Na^+$ ion is, 4.5 M

6 0

2 years ago

Two different compounds are obtained by combining nitrogen with oxygen. The first compound results from combining 46.7 gg of NN

mina [271]

Answer:

4.02

Explanation:

The mass ratios will be given by dividing the mass of O₂ into the mass of N₂.

So lets do our calculations:

First Compound:

53.3 g O₂ / 46.7 g N₂ = 1.14

Second Compound:

82.0 g O₂ / 17.9 g N₂ = 4.58

Ratio = 4.58 / 1.14 = 4.02

This result for all practical purposes is a whole number, and it is telling us that there are 4 times as many oxygen atoms in the second coumpound as in the first compound. This is so because the ratio we just calculated is also the ratio in mol atoms:

Ratio = [ mass O₂ / MW O2/ mass N₂/ MW N₂] 2nd compound / [mass O₂ / MW O2/ mass N₂/ MW N₂ !st compound]

and the molecular weights cancel each other.

The only N and O compounds that follow this ratio are N₂O₄ and N₂O, and this question could be made in a multiple choice to match formulas.

3 0

2 years ago