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Mashutka [201]
2 years ago
12

Which element is not an alkali metal?

Chemistry
2 answers:
irinina [24]2 years ago
5 0

Answer:

hydrogen, lithium, sodium element are not alkali metal

jasenka [17]2 years ago
4 0

Hello!

Answer:

Hydrogen

Explanation:

Hydrogen (H) has one electron in its outermost shell, but it is not classed as an alkali metal since it is not a metal but a gas at room temperature.

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A chemistry student needs 45.0mL of pentane for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the stud
Masja [62]

The mass of pentane the student should weigh out is

The density of pentane is 0.626 gcm-3

To calculate the mass of pentane following expression is used,

(Density is defined as the mass divide by volume)

Density = mass / volume

mass of pentane = Density of pentane * Volume of pentane

mass of pentane = 0.626 gcm-3 * 45.0 mL

                             = 28.17 g

Here the unit of mass of pentane is g,

However the unit of density is gcm-3 and unit of volume is mL i.e. cm3

Hence,   Mass = gcm-3 * cm3

              Mass = g

The mass of pentane the student should weigh out is 28.17g

Learn more about Density on

brainly.com/question/1354972

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5 0
2 years ago
Can someone answer 1-10 for me or the one you know I’ll give you brainlest!!!
Reika [66]
1. Coefficient
2. Subscript
3. Products
4. Reactants
5. Balance
6. Combustion
7. Decomposition
8. Single replacement
9. Double replacement
10. Synthesis
I’m sorry if I’m wrong
5 0
3 years ago
4. Consider the following half-reactions: MnO4–(aq) + 8H+(aq) + 5e– → Mn+2(aq) + 4H2O(l) NO3–(aq) + 4H+(aq) + 3e– → NO(g) + 2H2O
jeyben [28]

Explanation:

The chemical reaction given in the question is as follows -

MnO₄⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn²⁺ (aq) + 4H₂O (l)

NO₃⁻ (aq) + 4H⁺ (aq) + 3e⁻ → NO (g) + 2H₂O (l)

As we know , the value for reduction potential are -

Mn²⁺ = + 1.51  V

NO₃⁻  =  +0.96 V

From , the data given above , the value of the reduction potential of NO₃⁻ is less than the reduction potential of Mn²⁺ .

Hence ,

NO₃⁻  can not oxidize Mn²⁺ .

5 0
3 years ago
How many grams of hydrogen chloride can be produced from 1g of hydrogen and 55g of chlorine? What is the limiting reactant?
vova2212 [387]

Answer:

The limiting reactant is hydrogen, and the grams HCl produced is 36.175 g.

Explanation:

Balanced equation is 2 H + Cl2 = 2 HCl.

First thing, convert grams to moles via using molar mass.

Molar mass for hydrogen is 1.0079 g/mol. 1g x 1 mol / 1.0079 g = 0.99216 mol.

Molar mass for chlorine is 70.906 g/mol. 55g x 1 mol / 70.906 g = 0.7756748 mol.

Next, determine which is the limiting reactant - probably the fastest way to do it is just to take one of the reactants, say it's the limiting one, and calculate how much of the other reactant would be needed if that really was the limiting reactant, and then compare it to the actual moles of reactant available.

If hydrogen was the limiting reactant at 0.992 mol, you'd need .496 mol of Cl2 to complete the reaction.

If chloride was the limiting reactant at 0.776 mol, you'd need 1.55 mol of H to complete the reaction.

Comparing these numbers to the amounts we actually have available, the limiting reactant is hydrogen.

Once you've determined that, just plug in the amounts to the balanced equation to get the number of moles of HCL produced, which in this case, is just 0.992 mol.

Now, reverse the process that you took to get the moles of reactant, and you have the grams of product produced.

0.992 mol x 36.4609 g / 1 mol = 36.175 g.

7 0
3 years ago
The following question appears on a quiz: ""You fill a tank with gas at 60°C to 100 kPa and seal it. You decrease the temperatu
dusya [7]

Answer: The final pressure will decrease ad the value is 85 kPa

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=100kPa\\T_1=60^0C=(60+273)K=333K\\P_2=?\\T_2=10^0C=(10+273)K=283K

Putting values in above equation, we get:

\frac{100kPa}{333K}=\frac{P_2}{283K}\\\\P_2=85kPa

Hence, the final pressure will decrease ad the value is 85 kPa

8 0
3 years ago
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