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3 years ago

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Express250,000ml in liters

1 answer:

frutty [35]3 years ago

7 0

Answer:

250,000 ML is equal to 250 Liters.

Explanation:

The term 'Mili' multiplies the term 'Liter' by 1,000. Per every liter is 1000 milliliters.

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Write the 1 g equivalents using the following prefixes: (a) mega−, (b) kilo−, (c) deci−, (d) centi−, (e) milli−, (f) micro−, (g)

dem82 [27]

<u>Explanation:</u>

Unit is defined as the quantity that is used as a standard for measurement.

S.I. unit of mass of kilograms. Other units in which mass of a substance can be measured are grams, milligrams, etc...

All these units are inter convertible.

<u>For a:</u> Mega-

The conversion factor used is:

$1g=10^{-6}Mg$

<u>For b:</u> kilo-

The conversion factor used is:

$1g=10^{-3}kg$

<u>For c:</u> deci-

The conversion factor used is:

$1g=10dg$

<u>For d:</u> centi-

The conversion factor used is:

$1g=10^{3}cg$

<u>For e:</u> milli-

The conversion factor used is:

$1g=10^3mg$

<u>For f:</u> micro-

The conversion factor used is:

$1g=10^6\mu g$

<u>For g:</u> nano-

The conversion factor used is:

$1g=10^9ng$

<u>For h:</u> pico-

The conversion factor used is:

$1g=10^{12}pg$

8 0

3 years ago

If the sample has a total mass of 5.76 g and contains 1.79 g k, what are the percentages of kbr and ki in the sample by mass

Luden [163]

The percentages of KBr and KI in the sample by mass is 80.68 and 19.32 % respectively.

<h3>What is Molar Mass ?</h3>

Molar mass is defined as the mass contained in 1 mole of sample.

It is given that

the sample has a total mass of 5.76 g contains KBr and KI

and contains 1.79 gm of K is present

what is the percentage of KBr and KI in the sample

Molecular weight of K = 39

Molecular weight of Br = 79.9

Molecular weight of I = 126.9

In KBr the mass percentage of K is 39/(39+79.9) = 32.89%

In KI the mass percentage of K is 39/(39+126.9) = 23.5%

Let the mass of KBr present in the sample is x

K will be 0.3289 x

and let the mass of KI present be y

K will be 0.235y

x +y =5.76

0.3289x+0.235y = 1.79

0.0939y = 0.1045

y = 1.1125 gm

x = 5.76-1.1125

x = 4.6475 gm

% of KBr = (4.6475/5.76 )*100 = 80.68 %

% of KI = (1.1125/5.76) *100 = 19.32%

To know more about Molar Mass

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8 0

3 years ago

Holly has an unknown substance in a beaker. She wants to determine the relative pH of the unknown substance. She places a piece

Taya2010 [7]

Answer:

The correct answer is B has a neutral pH

Explanation:

we all know that acid converts blue litmus to red on the other hand base converts red litmus paper to blue.

According to the question the blue litmus paper does not change its colour or it stays blue that means the pH of solution is neutral or7 which prevent the colour change of the litmus paper.

If the solution was acidic then the blue litmus paper changes into red .

5 0

3 years ago

A sample of a compound contains only the elements sodium, sulfur, and oxygen. it is found by analysis to contain 0.979 g na, 1.3

Aloiza [94]

To determine the empirical formula for the compound that contains <span>0.979 g Na, 1.365 g S, and 1.021 g O, we convert these to mole units. The molar masses to be used are:

Molar mass of Na = 23 g/mol
</span>Molar mass of S = 32 g/mol
Molar mass of O = 16 g/ mol

The number of moles is obtained using the molar mass for each element.

moles Na = 0.979 g Na/ 23 g/mol Na = 0.04256
moles S = 1.365 g Na/ 32 g/mol Na = 0.04265
moles O = 1.021 g O/ 16 g/mol Na = 0.06326

We then divide each with the smallest number of moles obtained.

Na: 0.04256/ 0.04256 = 1
S: 0.04265/ 0.04256 = 1.002 ≈ 1
O: 0.06326/ 0.04256 = 1.49 ≈ 1.5

We then have an empirical formula of NaSO₁.₅. However, chemical formulas must have only integers as subscripts, thus, we multiply each to 2. The empirical formula is then Na₂S₂O₃ also known as sodium thiosulfate.

4 0

3 years ago

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Being able to roll the tongue is dominant over not being able to roll the tongue.

-Dominant- [34]

Answer:

Its 0

Explanation:

7 0

3 years ago

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