All categories

3 years ago

You have been given two unmarked containers.

Chemistry

1 answer:

insens350 [35]3 years ago

6 0

We need the full question to answer sorry

You might be interested in

lead can react with oxygen gas. If lead (IV) oxide is the product of the reaction, how would the reaction be classified

slamgirl [31]

Answer:

this reaction is an oxidation reaction

7 0

3 years ago

If I initially have 4.0 L of a gas at a pressure of 1.1 atm and a temperature of 298 K, what will the volume be if I increase th

DochEvi [55]

Answer:

1.34L

Explanation:

1 torr = 0.00132atm

P2= 2584torr = 3.41atm

T2= 37°C = 273+37 = 310K

Using combined gas law

P1V1/T1 = P2V2/T2

(1.1×4.0)/298 = (3.41 × V2)/310

V2= (1.1×4×310)/(298×3.41)

V2= 1.34L

4 0

3 years ago

Does anyone know the answer to this

Damm [24]

It’s Tetracarbon dioxide :)

6 0

3 years ago

What volume is occupied by 0.109 molmol of helium gas at a pressure of 0.98 atmatm and a temperature of 307 K

KATRIN_1 [288]

Answer:

2.8 L

Explanation:

From the question given above, the following data were obtained:

Number of mole (n) = 0.109 mole

Pressure (P) = 0.98 atm

Temperature (T) = 307 K

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

The volume of the helium gas can be obtained by using the ideal gas equation as follow:

PV = nRT

0.98 × V = 0.109 × 0.0821 × 307

0.98 × V = 2.7473123

Divide both side by 0.98

V = 2.7473123 / 0.98

V = 2.8 L

Thus, the volume of the helium gas is 2.8 L.

6 0

3 years ago

A 250. ML sample of 0.3M HCl is partially neutralized by the addition of 100. ML of 0.25M NaOH.Find the concentration of hydroch

Dahasolnce [82]

Answer:- 0.143 M

Solution:- HCl and NaOH reacts in 1:1 mol ratio as shown in the below reaction:

$HCl(aq)+NaOH(aq)\rightleftharpoons NaCl(aq)+H_2O(l)$

Let's calculate the initial moles of HCl and the moles of NaOH added to it:

$250.mL(\frac{1L}{1000mL})(\frac{0.3molHCl}{1L})$

= 0.075 mol HCl

$100.mL(\frac{1L}{1000mL})(\frac{0.25molNaOH}{1L})$

= 0.025 mol NaOH

Since they react in 1:1 mol ratio, 0.025 mol of NaOH will react with 0.025 moles of HCl.

Remaining moles of HCl = 0.075 - 0.025 = 0.050

Total volume of the resulting solution = 0.250 L + 0.100 L = 0.350 L

So, the concentration of HCl in the resulting solution = $\frac{0.050mol}{0.350L}$

= 0.143 M

Hence, the concentration of HCl acid in the resulting solution is 0.143 M.

6 0

4 years ago