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IRISSAK [1]
3 years ago
5

Solids can often diffuse into a liquid but a solid cannot diffuse into another child

Chemistry
1 answer:
Lady_Fox [76]3 years ago
4 0
Solid*? this is not a question.
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What is the mass in 6.34x10^23 molecules of calcium chloride
Liono4ka [1.6K]

Answer : The mass of calcium chloride is, 116.84 grams

Solution : Given,

Molar mass of calcium chloride, CaCl_2 = 110.98 g/mole

Number of molecules of calcium chloride = 6.34\times 10^{23}

As we know that,

1 mole of calcium chloride contains 6.022\times 10^{23} molecules of calcium chloride

or,

1 mole of calcium chloride contains 110.98 grams of calcium chloride

Or, we can say that

As, 6.022\times 10^{23} molecules of calcium chloride present in 110.98 grams of calcium chloride

So, 6.34\times 10^{23} molecules of calcium chloride present in \frac{6.34\times 10^{23}}{6.022\times 10^{23}}\times 110.98=116.84grams of calcium chloride

Therefore, the mass of calcium chloride is, 116.84 grams

6 0
3 years ago
What is the volume of 11.2 g of O2 at 7.78 atm and 415 K?
irina1246 [14]

Answer:

1.53 L

Explanation:

Step 1: Given data

  • Mass of oxygen (m): 11.2 g
  • Pressure (P): 7.78 atm
  • Temperature (T): 415 K
  • Ideal gas constant (R): 0.0821 atm.L/mol.K

Step 2: Calculate the moles (n) corresponding to 11.2 g of oxygen

The molar mass of oxygen is 32.00 g/mol.

11.2 g × (1 mol/32.00 g) = 0.350 mol

Step 3: Calculate the volume of oxygen

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T / P

V = 0.350 mol × (0.0821 atm.L/mol.K) × 415 K / 7.78 atm

V = 1.53 L

3 0
3 years ago
Find the volume of a rectangular prism that is<br> 25 cm by 10 cm by 15 cm.<br> height<br> length
Stolb23 [73]

Answer:

3750 cm.

Explanation:

You multiply the three side measurements to find the volume.

25cm·10cm·15cm

375cm·10cm

3750 cm.

<em><u>Hope this helps!</u></em>

4 0
3 years ago
Read 2 more answers
Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

7 0
3 years ago
Why does a 4 carbon linear alkane, butane c4h10, have two more hydrogens ?
Scilla [17]
Alkanes are hydrocarbons that only contain single bonds in them. A carbon can bond with up to 4 atoms, even with another carbon atom. So, in a C-C bond, 3 more H atoms can bond to each of the C atom. Generally, the chemical formula for alkanes is CₓH₂ₓ₊₂. So for butane, there are 4 C atoms. The corresponding H atoms are 2(4) + 2 = 10. That's why it's chemical formula is C₄H₁₀.
3 0
3 years ago
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