Who said the location and velocity of an electron cant be known

If 5.0 liters H2 (g) at STP is heated to a temperature of 985, pressure remaining constant, the new volume of the gas will be?

vlada-n [284]

Answer:

$V_2=18 \ L \ H_2$

General Formulas and Concepts:

Chemistry - Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
Charles' Law: $\frac{V_1}{T_1} =\frac{V_2}{T_2}$

Explanation:

Step 1: Define

Initial Volume: 5.0 L H₂ gas

Initial Temp: 273 K

Final Temp: 985 K

Final Volume: ?

Step 2: Solve for new volume

Substitute: $\frac{5 \ L \ H_2}{273 \ K} =\frac{x \ L \ H_2}{985 \ K}$
Cross-multiply: $(5 \ L \ H_2)(985 \ K) = (x \ L \ H_2)(273 \ K)$
Multiply: $4925 \ L \ H_2 \cdot K = 273x \ L \ H_2 \cdot K$
Isolate x: $18.0403 \ L \ H_2 = x$
Rewrite: $x=18.0403 \ L \ H_2$

Step 3: Check

We are given 2 sig figs as the smallest. Follow sig fig rules and round.

$18.0403 \ L \ H_2 \approx 18 \ L \ H_2$

5 0

2 years ago

What is the daughter nucleus of pt-199 after a beta decay?

Setler79 [48]

B. At-200

In beta decay the atomic number increases by 1

6 0

2 years ago

If 1.85 g of Mg(OH)2 reacts with 3.71 g of HCl,

Damm [24]

Both of them are a hope this helps

8 0

2 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

2 years ago

Carbon dioxide is released into the atmosphere by _____.

kirill115 [55]

Answer:

4 a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation.;

Explanation:

a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation

3 0

2 years ago