2 years ago

How should you measure the mass of a given powdered substance??

1 answer:

6 0

To measure the mass of a given powdered substance do the following:
1. Zero the weighing balance, that is, put the weigh balance at zero mark.
2. Put an empty weighing paper on the balance and take its reading. Record it as X1.
3. Use spatula to put the powder you want to measure on the weighing paper on the balance. Take your reading and record it as X2.
The mass of the powered is X2 - X1.

You might be interested in

In order to wash out the impurity, you looked up some sketchy method from 1952 which Hank Pym wrote during his Ph. D. under E. J

Eva8 [605]

Answer:

516.24

Explanation:

The solved solution is in the attached document.

8 0

2 years ago

Who helps recycle nutrients through an ecosystem? A. producers and decomposers B. producers and consumers C. consumers and decom

s2008m [1.1K]

D. Producers im sure of it hope im right

8 0

3 years ago

Read 2 more answers

What is called mother liquor in a chemistry lab?

Paladinen [302]

The mother liquor<span> is the part of a solution that is left over after it crystallizes. An example of a place you could find this is in sugar refinement.</span>

7 0

3 years ago

Which inorganic substance taken up by plants is also taken in and used directly by animals, including humans?

masha68 [24]

Answer: water

water can not be organic...but is is took by humans and plants because theur life dpends on it

Explanation:

3 0

3 years ago

Read 2 more answers

A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0

dimaraw [331]

Explanation:

It is known that for high concentration of $M^{2+}$ , reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.

Now, $E^{o}_{cell}$ = 0 and the general reaction equation is as follows.

$M^{2+} + M \rightarrow M + M^{2+}$

3.00 M n = 2 30 mM

E = $0 - \frac{0.0591}{2} log \frac{50 \times 10^{-3}}{1}$

= $-\frac{0.0591}{2} log (5 \times 10^{-2})$

= 0.038 V

Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.

5 0

3 years ago