The balanced equation for the combustion of ethene is given below. If 97.7 g of oxygen is used, how much

If this atom has a balanced charge, how many protons would you expect to find in this atom?

uranmaximum [27]

The number of protons would be equivalent to the number of electrons if the net charge on the atom is 0.

6 0

3 years ago

What mass of lead (II) chloride is produced when 200.0 mL of a 0.250 M solution of sodium chloride is mixed with 200.0 mL of a 0

kow [346]

Answer:

Option d. 6.95 g

Explanation:

First of all, we state the reaction:

2NaCl + Pb(NO₃)₂ → PbCl₂ + 2NaNO₃

We determine the moles of each reactant, to state the limiting

Firstly we convert volume frm mL to L

0.200 L . 0.250M = 0.05 moles of NaCl

0.200L . 0.250M = 0.05 moles of Pb(NO₃)₂

Acording to stoichiometry we know that relation is 1:2, so the limiting reagent is the NaCl.

For 1 mol of Pb(NO₃)₂ I need 2 moles of NaCl

For 0.05 moles of Pb(NO₃)₂ I would need, the double → 0.1 moles

(We only have, 0.05 moles of NaCl)

Stoichiometry to the formed product is 2:1

From 2 moles of NaCl I produce 1 mol of PbCl₂

From 0.05 moles I would produce, the half → 0.025 moles

Let's convert the moles to mass → 0.025 mol . 278.1 g / 1mol = 6.95 g

8 0

3 years ago

A person tries to heat up her bath water by adding 5.0 l of water at 80c to 60 l of water at 30c. what is the final temperature

Sophie [7]

We can calculate the final temperature from this formula :

when Tf = (V1* T1) +(V2* T2) / (V1+ V2)

when V1 is the first volume of water = 5 L

and V2 is the second volume of water = 60 L

and T1 is the first temperature of water in Kelvin = 80 °C +273 = 353 K

and T2 is the second temperature of water in Kelvin = 30°C + 273= 303 K

and Tf is the final temperature of water in Kelvin

so, by substitution:

Tf = (5 L * 353 K ) + ( 60 L * 303 K) / ( 5 L + 60 L)

= 1765 + 18180 / 65 L

= 306 K
= 306 -273 = 33° C

8 0

3 years ago

Anybody help?! please

Mashutka [201]

J.j Thomson electrons demo true indivisible sphere n Ernest central nucleas

3 0

3 years ago

A container initially holds 5.67 x 10^-2 mol of propane and has a volume of V1. The volume of the container was increased by add

Brut [27]

Answer:

Initial volume of the container (V1) = 1.27 L (Approx)

Explanation:

Given:

Number of mol (n1) = 5.67 x 10⁻²

Number of mol (n2) = (5.67 +2.95) x 10⁻² = 8.62 x 10⁻²

New volume (V2) = 1.93 L

Find:

Initial volume of the container (V1)

Computation:

Using Avogadro's law

V1 / n1 = V2 / n2

V1 / 5.67 x 10⁻² = 1.93 / 8.62 x 10⁻²

V1 = 10.9431 / 8.62

Initial volume of the container (V1) = 1.2695

Initial volume of the container (V1) = 1.27 L (Approx)

5 0

3 years ago