All categories

3 years ago

Which is the atomic number of an alkali metal? 10 11 12 13

Chemistry

2 answers:

prohojiy [21]3 years ago

5 0

Answer:

11 is the atomic number of alkali metal

Explanation:

The Group 1 elements are called the alkali metals. ... All the Group 1 elements are very reactive . They must be stored under oil to keep air and water away from them. Group 1 elements form alkaline solutions when they react with water, which is why they are called alkali metals.

kvv77 [185]3 years ago

3 0

I think it’s 11 for the alkali metal

You might be interested in

This is actually science but anyways my teacher didn't explain much can someone help?

zepelin [54]

TT homo
Tt hetero
tt homo

5 0

3 years ago

What is the mass in grams of 85.32 mL of blood plasma with a density of 1.03 g/mL

sveticcg [70]

Well if you're looking for grams, all you need to do is cancel out units. (ml)(g/ml)=g because the ml cancels out. Thus, multiply: (85.32ml)(1.03g/ml)=...I'll let you solve this. :) Good luck! Hope that helped. When in doubt, look at the units.

3 0

3 years ago

Is Helium a compound,element, or.mixture

dusya [7]

Answer:

element

Explanation:

we know that helium is a pure substance although helium atoms are sometimes mixed with their isotopes it is still the same element. since there is no other element combined with helium this makes it an element.

5 0

3 years ago

A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path

Igoryamba

Explanation:

For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

R = 8.314 J/(K mol), T = 298 K ,

$V_{2}$ = 8.34 L, $V_{1}$ = 2.67 L

Now, putting the given values into the above formula as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

= -2818.68 J

Hence, work for path A is -2818.68 J.

For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure $P_{external}$ = 1.00 atm.

So, W = $-P_{external} \times \Delta V$

Now, putting the given values into the above formula as follows.

W = $-P_{external} \times \Delta V$

= $-1 atm \times (8.34 L - 2.67 L)$

= -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

= -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

W = -574.54 J

Hence, work for path B is -574.54 J.

3 0

4 years ago

Suppose a grill lighter contains 50.0 g of butane. How many grams of butane in the lighter would have to be burned to produce 17

Hitman42 [59]

Answer is: mass of burned butane is 11.6 g.

Chemical reaction: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.

m(butane) = 50,0 g.

V(CO₂) = 17,9 L.
n(CO₂) = V(CO₂) ÷ Vm.
n(CO₂) = 17,9 L ÷ 22,4 L/mol.
n(CO₂) = 0,8 mol.
From chemical reaction n(CO₂) : n(C₄H₁₀) = 8 : 2.
n(C₄H₁₀) = 0,8 mol ÷ 4.
n(C₄H₁₀) = 0,2 mol.
m(C₄H₁₀) = n(C₄H₁₀) · M(C₄H₁₀).
m(C₄H₁₀) = 0,2 mol · 58 g/mol.
m(C₄H₁₀) = 11,6 g.

5 0

3 years ago