How many moles of solute are contained in 3.0 L of a 1.5 M solution? Helppp pleaseee

Poor precision in scientific measurement may arise from

Arturiano [62]

Based on the experience of the responder, to correctly calculate measurements in real-world. Firstly is to avoid errors as much as possible. Errors are what makes your measurement invalid and unreliable. There are two types of error which is called the systematic error and the random error. Each error has different sources. Words that were mentioned –invalid and unreliable are very important key aspects to determine that your measure is truly accurate and consistent. Some would recommend using the mean method, doing three trials in measuring and getting their mean, in response to this problem.

7 0

4 years ago

When should a lab coat, safety goggles, and gloves be worn in the laboratory?

Sever21 [200]

I would say that you should wear a lab coat, safety goggles, and gloves when the teacher says so - not everything in a lab is dangerous, so there is no need to always wear these. But when the teacher says you should - then you should.

8 0

4 years ago

Read 2 more answers

The reaction A(B) = 2B(g) has an equilibrium constant of K = 0.045. What is the equilibrium constant for the reaction B(g) =1/2A

Mamont248 [21]

Answer:

The $K_c$ for the reaction $B(g) = \frac{1}{2}A$ will be 4.69.

Explanation:

The given equation is A(B) = 2B(g)

to evaluate equilibrium constant for $B(g) = \frac{1}{2}A$

$K_c=[B]^2[A]$

= 0.045

The reverse will be $2B\leftrightharpoons A$

Then, $K_c = \frac{[A]}{[B]^2}$

= $\frac{1}{0.045}$

= $22m^{-1}$

The equilibrium constant for $B(g) = \frac{1}{2}A$ will be

$K_c = \sqrt{K_c}$

$=\sqrt{22}$

= 4.69

Therefore, $K_c$ for the reaction $B(g) = \frac{1}{2}A$ will be 4.69.

5 0

3 years ago

A 29.7 g sample of iron ore is treated as follows. The iron in the sample is all converted by a series of chemical reactions to

Softa [21]

Answer: The mass of iron in the ore is 10.9 g

Explanation:

We are given:

Mass of iron (III) oxide = 15.6 g

We know that:

Molar mass of Iron (III) oxide = 159.69 g/mol

Molar mass of iron atom = 55.85 g/mol

As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.

To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:

In 159.69 g of iron (III) oxide, mass of iron present is $(2\times 55.85)=111.7g$

So, in 15.6 g of iron (III) oxide, mass of iron present will be = $\frac{111.7g}{159.69g}\times 15.6g=10.9g$

Hence, the mass of iron in the ore is 10.9 g

7 0

3 years ago

A student mixed two chemicals to allow them to

Westkost [7]

Answer:

B .it is an exothermic reaction

Explanation:

i need brainleyest

4 0

2 years ago