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3 years ago

Why are standards needed for measured quantities?

Chemistry

2 answers:

aleksandr82 [10.1K]3 years ago

8 0

It's imperative to have standard units for measuring quantities with the goal that the quantities are uniform.

Further Explanation:

measuring quantities:

In a physical setting an estimation instrument might be checked to estimating substances of a particular physical amount. In such a setting the particular physical amount is known as a deliberate amount. The synonymous thought "noticeable" regularly is utilized with regards to quantum mechanics.

Standard units:

which means of standard unit alludes to a unit of specific arrangement of units of estimation called the standard framework (versus the decimal measuring standard). The standard framework incorporates the standard units of the foot, the pound (mass), and the gallon.

Standard units are utilized in estimation:

A standard unit of estimation is a quantifiable language that enables everybody to comprehend the relationship of the article with the estimation. It is communicated in inches, feet, and pounds, in the United States, and centimeters, meters, and kilograms in the decimal standard for measuring.

Techniques for Measurement:

Direct Method of Measurement, In this technique for estimation, the obscure amount is straightforwardly contrasted and the standard amount. The consequence of the amount is communicated in number. It is the most widely recognized strategy for estimating the physical amounts like length, temperature, weight, and so on.

Subject: chemistry

Level: High School

Keywords: measuring quantities, Standard units, Standard units are utilized in estimation, Techniques for Measurement.

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Alex3 years ago

4 0

In chemistry the need for standards when measuring quantities are in place for accurate measurement that is recognized world wide. This way, no matter what language people use, they will always understand there standards.

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The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha

zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

$T = x*P$

$25.9 = x*10.0$

$25.9 = 10.0\:x$

$10.0\:x = 25.9$

$x = \dfrac{25.9}{10.0}$

$\boxed{x = 2.59}$

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

$m = \dfrac{m_o}{2^x}$

$m = \dfrac{53.3}{2^{2.59}}$

$m \approx \dfrac{53.3}{6.021}$

$\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

3 0

3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

3 years ago

How old is a bone if it now .3125 of C-14 when it originally had 80.0g of C-14

Taya2010 [7]

<h3>Answer:</h3>

42960 years

<h3>Explanation:</h3>

<u>We are given;</u>

Remaining mass of C-14 in a bone is 0.3125 g
Original mass of C-14 on the bone is 80.0 g
Half life of C-14 is 5370 years

We are required to determine the age of the bone;

Using the formula;

Remaining mass = Original mass × 0.5^n , where n is the number of half lives.

Therefore;

0.3125 g = 80.0 g × 0.5^n

3.90625 × 10^-3 = 0.5^n

Introducing logarithm on both sides;

log 3.90625 × 10^-3 = n log 0.5

Solving for n

n = log 3.90625 × 10^-3 ÷ log 0.5

= 8

Therefore, the number of half lives is 8
But, 1 half life is 5370 years
Therefore;

Age of the rock = 5370 years × 8

= 42960 years

Thus, the bone is 42960 years old

7 0

3 years ago