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3 years ago

What is the final step in the scientific method

2 answers:

8 0

The final step in the scientific method is the conclusion. The conclusion will either clearly support the hypothesis or it will not. If the results support the hypothesis a conclusion can be written

ehidna [41]3 years ago

5 0

It’s the conclusion which is the result

You might be interested in

What is the net ionic equation for the reaction of solid iron with aqueous copper sulfate?

lozanna [386]

This is a single replacement reaction. So the final products are a metal and a salt with replaced cation. In this case, the final products are copper (metal) and iron sulfate (salt).

8 0

3 years ago

How many of these transition metals have you heard of before?

AleksAgata [21]

Answer:

38 elements or Element 31: Gallium Element 32: Germanium Element 33: Arsenic Element 34: Selenium Element 35: Bromine Element .

Explanation:

How many transition metals are there?

The 38 elements in groups 3 through 12 of the periodic table are called "transition metals". As with all metals, the transition elements are both ductile and malleable, and conduct electricity and heat.

6 0

2 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

4 0

3 years ago

Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne

777dan777 [17]

Answer:

Explanation:

From the information given:

$CaF_2 \to Ca^{2+} + 2F^-$

$Ksp = 3.2 \times 10^{-11}$

no of moles of $Ca^{2+}$ = 0.01 L × 0.0010 mol/L

no of moles of $Ca^{2+}$ = $1 \times 10^{-5} \ mol$

no of moles of $F^-$ = 0.01 L × 0.00010 mol/L

no of moles of $F^-$ = $1 \times 10^{-6}\ mol$

Total volume = 0.02 L

$[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L$

$[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}$

$[F^{-}] = 5 \times 10^{-5} \ mol/L$

$Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}$

Since Q<ksp, then there will no be any precipitation of CaF2

3 0

3 years ago

If the sun is next to the earth on the left, and the moon is next to the earth on the right, what type of tide is it?

Andrej [43]

Answer:

When the sun, moon, and Earth are in alignment (at the time of the new or full moon), the solar tide has an additive effect on the lunar tide, creating extra-high high tides, and very low, low tides—both commonly called spring tides.

3 0

2 years ago