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2 years ago

6

Newton’s first law of motion states that an object at rest stays at rest,and an object in motion stays in motion,unless acted on

by an unbalanced force.

1 answer:

julia-pushkina [17]2 years ago

7 0

Answer: yes that is true

Explanation:

newtons first law does state that.

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Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution for concentrations up to appro

Hatshy [7]

Answer:

The unit cell edge length for the alloy is 0.288 nm

Explanation:

Given;

concentration of vanadium, Cv = 8 wt%

concentration of iron, Cfe = 92 wt%

density of vanadium = 6.11 g/cm³

density of iron = 7.86 g/cm³

atomic weight of vanadium, Av = 50.94 g/mol

atomic weight of iron, Afe= 55.85 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_v}{\rho _v} + \frac{C__{Fe}}{\rho _{Fe}} }$

$\rho _{Avg.} = \frac{100}{\frac{8}{6.11} + \frac{92}{7.86} } = 7.68 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _v} + \frac{C__{Fe}}{A _{Fe}} }$

$A _{Avg.} = \frac{100}{\frac{8}{50.94} + \frac{92}{55.85} } = 55.42 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a BCC crystal structure, there are 2 atoms per unit cell; n = 2

$V_c=\frac{2*55.42}{ 7.68*6.022*10^{23}} = 2.397*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (2.397*10^{-23}}){^\frac{1}{3}}\\\\a = 2.88 *10^{-8} \ cm$ = 0.288 nm

Therefore, the unit cell edge length for the alloy is 0.288 nm

4 0

3 years ago

A chemical engineer studying the properties of fuels placed 1.670 g of a hydrocarbon in the bomb of a calorimeter and filled it

tigry1 [53]

Answer:

856,68 J/gram

Explanation:

We know that the variaton of temperature of 1K is equal to 1°C

so if the ΔT is 3.55°C, it is 3.55K.

so

403 J - 1 K

x - 3.55K

X= 403 * 3.55 = 1430,65 J

1.670 g - 1430,65 J

1 g - y

y = $\frac{1430,65}{1.670}$ = 856,68 J

7 0

3 years ago

_Al(NO3)3 +_(NH4)3 P04_AlPO4+_ NH, NO3<br>balancing equations

astra-53 [7]

Answer:

Al(NO₃)₃ + (NH₄)₃PO₄ —> AlPO₄ + 3NH₄NO₃

The coefficients are: 1, 1, 1, 3

Explanation:

__Al(NO₃)₃ + __(NH₄)₃PO₄ —> __AlPO₄ + __NH₄NO₃

The above equation can be balance as illustrated below:

Al(NO₃)₃ + (NH₄)₃PO₄ —> AlPO₄ + NH₄NO₃

There are 12 atoms of H on the left side and 4 atoms on the right side. It can be balance by writing 3 before NH₄NO₃ as shown below:

Al(NO₃)₃ + (NH₄)₃PO₄ —> AlPO₄ + 3NH₄NO₃

Now the equation is balanced.

The coefficients are: 1, 1, 1, 3

4 0

2 years ago

A reaction between an acid and a base is called a

Sergeeva-Olga [200]

Neutralization reaction

5 0

3 years ago

Why is dissolving salt in water is a physical change? Help me please

svlad2 [7]

it is a great change because the salt goes in and turn the water to salt water

7 0

3 years ago

Read 2 more answers