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3 years ago

When copper (II) chloride reacts with sodium nitrate, copper (II) nitrate and sodium chloride are formed.

Chemistry

1 answer:

anzhelika [568]3 years ago

3 0

Answer:

How many grams of copper (II) nitrate is formed

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The diameter of carbon atom is 0.000000000 154m. What is the number expressed in scientific notation

kogti [31]

1.54×10 −10

one and fifty four-hundreths times ten to the power of negitiive 10

7 0

3 years ago

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When substances are evenly spread throughout a mixture, it is called ____________________?

harina [27]

The answer is _Dissolving_.

Hope tis helps

5 0

3 years ago

Suppose a 10.0 mL sample of an unknown

mote1985 [20]

The concentration of HCl is equal to 2.54mol/L.

<h3>Mole calculation</h3>

To solve this question, one must use the molarity calculation, which corresponds to the following expression:

$M = \frac{mol}{v}$

Thus, to find the molarity of the sample, the following calculations must be performed:

$V_f = 10ml + 625ml = > 635ml$

$\frac{0.004mol}{xmol} =\frac{1000ml}{635ml}$

$x = 0.00254 mol$

So, 0.00254 moles were added per 10ml, so we can do:

$\frac{0.00254mol}{xmol}= \frac{10ml}{1000ml} \\x = 2.54mol/L$

So, the concentration of HCl is equal to 2.54mol/L.

Learn more about mole calculation in: brainly.com/question/2845237

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

The pathogen below is a single-celled organism without a nucleus that can cause illness in a humans.

sukhopar [10]

Answer:

bacteria

Explanation:

The pathogen below is a single-celled organism without a nucleus that can cause illness in a humans.

3 0

2 years ago