( BRAINLIEST) Name the Salt formed by Sulfuric acid

Determine the mass of water produced from 50g hydrochloric acid and

Alenkinab [10]

Answer:

24.7 grams H₂O

Explanation:

To find the mass of water, you should (1) convert from grams HCl to moles (via molar mass from periodic table), then (2) convert from moles HCl to moles H₂O (via mole-to-mole ratio from equation), and then (3) convert from moles H₂O to grams (via molar mass from periodic table).

1 HCl + KOH --> 1 H₂O + KCl

Molar Mass (HCl): 1.008 g/mol + 35.45 g/mol
Molar Mass (HCl): 36.458 g/mol

Molar Mass (H₂O): 2(1.008 g/mol) + 16.00 g/mol
Molar Mass (H₂O): 18.016 g/mol

50 g HCl 1 mole HCl 1 mole H₂O 18.016 g
-------------- x ------------------ x ------------------- x ------------------- = 24.7 g H₂O
36.458 g 1 mole HCl 1 mole H₂O

3 0

2 years ago

Compare uniformitarianism with catastrophism

wolverine [178]

Catastrophism is a theory that stated that earth was shaped by a suddentmassive violent events that happened within a short period of time.
Uniformitarianism, on the other hand, is a theory that stated earth was shaped by slow and gradual events over a long period of time, which could take millions of years.

3 0

3 years ago

Igneous rock is formed by the cooling and solidification of molten Earth materials. Igneous rock forms when magma or lava cools.

konstantin123 [22]

First blank = cools. Not sure about the second blank.

Hope this helped at least a bit :)

6 0

3 years ago

Read 2 more answers

A chemist wants to find Kc for the following reaction at 751 K: 2NH3(g) + 3 I2 (g) LaTeX: \Longleftrightarrow ⟺ 6HI(g) + N2(g) K

charle [14.2K]

Answer: The equilibrium constant for the total reaction is $4.09\times 10^{-6}$

Explanation:

We are given:

$K_{c_1}=0.282\\\\K_{c_2}=41$

We are given two intermediate equations:

Equation 1: $N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g);K_{c_1}=0.282$

The expression of $K_{c_1}$ for the above equation is:

$K_{c_1}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

$0.282=\frac{[NH_3]^2}{[N_2][H_2]^3}$ .......(1)

Equation 2: $H_2(g)+I_2(g)\rightleftharpoons 2HI(g);K_{c_2}=41$

The expression of $K_{c_2}$ for the above equation is:

$K_{c_2}=\frac{[HI]^2}{[H_2][I_2]}$

$41=\frac{[HI]^2}{[H_2][I_2]}$ ......(2)

Cubing both the sides of equation 2, because we need 3 moles of HI in the main expression if equilibrium constant.

$(41)^3=\frac{[HI]^6}{[H_2]^3[I_2]^3}$

Now, dividing expression 1 by expression 2, we get:

$\frac{K_{c_1}}{K_{c_2}}=\left(\frac{\frac{[NH_3]^2}{[N_2][H_2]^3}}{\frac{[HI]^6}{[H_2]^3[l_2]^3}}\right)\\\\\\\frac{0.282}{68921}=\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}$

$\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}=4.09\times 10^{-6}$

The above expression is the expression for equilibrium constant of the total equation, which is:

$2NH_3(g)+3I_2(g)\rightleftharpoons 6HI(g)+N_2;K_c$

Hence, the equilibrium constant for the total reaction is $4.09\times 10^{-6}$

8 0

3 years ago

Phosphorus has three electrons in its 3p sublevel and sulfur has four. Phosphorus should have the lower ionization energy but it

Nadya [2.5K]

Answer:

B

Explanation:

First of all it is important to know that a half filled orbital is particularly stable. In phosphorus all the electrons occur singly in the 3p sublevel minimizing inter electronic repulsion hence it is more difficult to remove an electron from this energetically stable arrangement. In sulphur, electrons are paired in one of the 3p orbitals thereby lowering the energy of that level due to instability caused by interelectronic repulsion between two electrons in the same orbital.

8 0

3 years ago