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Mars2501 [29]
3 years ago
5

The following characteristics describe which of these terms? the substance is not uniform in appearance and each part of the mix

ture contains a combination of different ingredients in different ratios.
A. homogeneous mixtures
B. heterogeneous mixtures
C. pure substance
D. solution
Chemistry
2 answers:
Mademuasel [1]3 years ago
6 0

B. heterogeneous mixtures

Katyanochek1 [597]3 years ago
3 0
B. Heterogeneous mixture
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No sorry because i'm only 12 years old so no

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3 years ago
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The heat of vaporization of water at 100°c is 40.66 kj/mol. Calculate the quantity of heat that is absorbed/released when 9.00 g
timofeeve [1]

Answer:

20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.

Explanation:

Heat is being consumed during vaporization and heat is being released during condensation.

To vaporize 1 mol of water, 40.66 kJ of heat is being consumed.

Molar mass of water = 18.02 g/mol

Hence, to vaporize 18.02 g of water , 40.66 kJ of heat is being consumed.

So, to vaporize 9.00 g of water, (\frac{40.66}{18.02}\times 9.00)kJ of heat or 20.3 kJ of heat is being consumed

As condensation is a reverse process of vaporization therefore 20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.

5 0
3 years ago
Different isotopes of the same element must have a different
slega [8]

Answer:

they must have same atomic number and different atomic mass

6 0
3 years ago
A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr
umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

5 0
3 years ago
Complete combustion of 5.90 g of a hydrocarbon produced 19.2 g of CO2 and 5.89 g of H2O. What is the empirical formula for the h
scZoUnD [109]
44g \ CO_{2} \ \ \ \rightarrow \ \ 12g \ C\\
19,2g \ CO_{2} \rightarrow \ \ \ x\\\\
x=\frac{19,2g*12g}{44g}\approx5,24g \ \ \ \Rightarrow \ \ n=\frac{5,24g}{12\frac{g}{mol}}\approx0,44mol\\\\\\
18g \ H_{2}O \ \ \ \rightarrow \ \ \ 2g \ H\\
5,89g \ H_{2}O \ \rightarrow \ \ y\\\\
y=\frac{5,89g*2g}{18g}\approx0,65g \ \ \ \ \Rightarrow \ \ n=\frac{0,65g}{1\frac{g}{mol}}=0,65mol\\\\\\
n_{C}:n_{H}=0,44:0,65\approx1:1\\\\
CH
4 0
3 years ago
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