Help! 7th Grade Math!

How many moles of silver chloride are produced from 15.0 moles of silver nitrate

Ghella [55]

15.0 moles of silver chloride are produced.

7 0

3 years ago

How do you calculate the number of valence electrons in an atom?

gavmur [86]

To find the valence electrons in an atom, identify what group the element is in. An element in group 1A has 1 valence electron. For example, Li is in group 1A, so that means it has one valence electron. If the element is in group 2A, then it has two valence electrons.

5 0

3 years ago

What is the kinetic energy of a 5 kg object moving at 7 m/s?

valina [46]

Answer:

122.5 Joule, or 122.5 j

Explanation:

Given,

mass of the object (m) = 5 kg

velocity of the object (v) = 7 m/s

Kinetic energy = $\frac{1}{2}$ × m × v²

Applying the formula:

Kinetic energy = $\frac{1}{2}$ × 5 × 7²

⇒ $\frac{1}{2}$ × 5 × 7 × 7

⇒ $\frac{245}{2}$

⇒ 122.5 Joule, or 122.5 j

Kinetic energy is the energy that an object gains as the result of the motion. It also depends on the mass of the object and force with which the motion is applied. In the given question, the mass of the object is 5 kg and the force of the velocity by which it is moving is 7 m/s.

3 0

3 years ago

The solid compound, K2SO4, contains?

givi [52]

The solid compound, K2SO4 contains a cation called K+ and an anion called SO42-. In this case, there are 2 atoms of potassium, 1 atom of sulfur and 4 moles of oxygen. The compound also contains ionic bonds because of the composing non-metals and metal.

3 0

3 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

3 years ago