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pantera1 [17]
3 years ago
10

The closeness of a measurement to its true value is a measure of its _____.

Chemistry
2 answers:
svlad2 [7]3 years ago
7 0
The closeness of a measurement to its true value is a measure of its accuracy. This term is the degree of which a certain measurement conforms to the correct value or the standard value. It is not the same with the term precision. Precision, on the other hand, is a measure used to characterize the closeness of the data measured.
Brilliant_brown [7]3 years ago
5 0

Answer: I just took the test and got 100%

1. B, a leaf changing color in autumn

2. B, well-tested explanation for a broad set of observations

3. C, evaluate

4. B, accuracy

5. C, 7.30 x 10^-7 km

6. B, 11mm^2

7. A. 2.9 x 10^7 m ^3

8. D, kelvin

9. B, 1dg

10. B, 0.987 g

11. B, 1.1 x 10^2 ml

12. C, decreases

13. D, -34*C

For the last 6 question please do not copy my exact answer

14. The chemistry that was before was based on observation. Thanks to him it is now based on measurement. He proved that in order for an object to burn, it requires oxygen for that process to occur.

15. Organic chemistry: study of carbon

Physical chemistry: study of chemistry with physics

Biochemistry: study of chemical process

Analytical chemistry: study and making of tools to measure matter

Inorganic chemistry: study of non organic compounds

16. Get creative :)

17. 225.7g

multiply the volume: 1*4*2.5=10

multiply the number above with the density of osmium

22.57*10=225.7g

18. It is easier to measure because it is in whole numbers(decimal) instead of fractions.

19. Mass is how much matter an object has. Weight is how much affect gravity has on that object. If I go to the moon, I will weight less because of the lack of gravity, but I will still retain the same mass.

Feel free to mark as brainliest :)

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‼️ i really need help. if anyone can do 10 pages of a grade 11 chemistry packet (multiple choice) let me know. i can pay money
Alja [10]

Answer:

mmmmmmmmmmmmmmmmmm

Explanation:

6 0
2 years ago
At what temperature will 0.654 moles of helium gas occupy 12.30 liters at 1.95 atmospheres?
DedPeter [7]

Answer:

447,25k

Explanation:

According to the ideal gas law

PV=nRT

Where:

P: is the pressure of the gas in atmospheres.

V: is the volume of the gas in liters.

n: number of moles of the gas

R: ideal gas constant

T: absolute temperature of the gas in kelvin

now using:

T=\frac{PV}{nR}\\ \\T=\frac{12,3L.1,95atm}{0,654mol.0,082(L.atm/K.mol)}\\ \\T=447,25K

3 0
3 years ago
Suppose that coal of density 1.5 g/cm^3 is pure carbon. (It is, in fact, much more complicated, but this is a reasonable first a
NISA [10]

Answer:

q = -6464.9 kJ

Explanation:

We are given that the heat of combustion is  ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.

vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³

m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g

mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol

q =  −394 kJ /mol C x 16.41 mol C = -6464.9 kJ

7 0
3 years ago
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
Which term refers to the loss of fertile soil from drying out?
dedylja [7]
Desertification.......
5 0
3 years ago
Read 2 more answers
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