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Ksju [112]
3 years ago
6

I just need the right answer plz

Chemistry
1 answer:
astraxan [27]3 years ago
5 0

Answer:

-0.5

Explanation:

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Recycling one ton of paper is the equivalent of approximately how much of the following? a.) all of the below b.) 460 gallons of
jek_recluse [69]
I think the answer is a
5 0
3 years ago
What was Ernest Rutherford's experiment?
Dafna1 [17]

Answer:

C. He shot tiny alpha particles through a piece of gold foil.

Explanation:

In the year 1911, Ernest Rutherford performed the gold foil experiment which gave a deeper perspective to the structure of an atom.

He simply collided a thin gold foil with an alpha particle which he generated from a radioactive source. He discovered that most of the alpha particles passed through the thin gold foil but a few were deflected back. His discovery led to the proposition of the nuclear model of the atom.

6 0
3 years ago
The molecular weight of a gas is ________ g/mol if 3.5 g of the gas occupies 2.1 l at stp
bija089 [108]
<span>Pre-1982 definition of STP: 37 g/mol Post-1982 definition of STP: 38 g/mol This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using. First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume. Pre-1982: 2.1 / 22.414 = 0.093691443 moles Post-1982: 2.1 / 22.71098 = 0.092466287 moles Now determine the molar mass. Simply divide the mass by the moles. So Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol Finally, round to 2 significant figures. So Pre-1982: 37 g/mol Post-1982: 38 g/mol</span>
5 0
4 years ago
What is the empirical formula for a compound that contains 10.89% magnesium 31.77% chloride and 57.34% oxygen
Blizzard [7]

Answer: Mg_{1}Cl_{2}O_{8}

Explanation:

If percentage are given then we are taking total mass is 100 grams.So, the mass of each element is equal to the percentage given.

Mass of Mg = 10.89 g

Mass of Cl = 31.77 g

Mass of O = 57.34 g

Step 1 : convert given masses into moles.

Moles of Mg=\frac {\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac {10.89g}{24g/mole}=0.45moles

Moles of Cl = \frac {\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac {31.77g}{35.5g/mole}=0.89moles

Moles of O =\frac {\text{ given mass of O}}{\text{ molar mass of O}}=\frac {57.34g}{16g/mole}=3.58moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Mg = \frac {0.45}{0.45}=1

For Cl = \frac {0.89}{0.45}=2

For O= \frac {3.58}{0.45}=8

The ratio of Mg :Cl : O= 1 : 2 : 8

Hence the empirical formula is Mg_{1}Cl_{2}O_{8}

4 0
3 years ago
Consider the reaction NOBr(g) =&gt; NO(g) + 1/2 Br2(g). A plot of 1/[NOBr] vs time give a straight line with a slope of 2.00 M-1
timama [110]

Answer:

Thus, the order of the reaction is 2.

The rate constant of the graph which is :- 2.00 M⁻¹s⁻¹

Explanation:

The kinetics of a reaction can be known graphically by plotting the concentration vs time experimental data on a sheet of graph.

The concentration vs time graph of zero order reactions is linear with negative slope.

The concentration vs time graph for a first order reactions is a exponential curve.  For first order kinetics the graph between the natural logarithm of the concentration vs time comes out to be a straight graph with negative slope.

The concentration vs time graph for a second order reaction is a hyberbolic curve. Also, for second order kinetics the graph between the reciprocal of the concentration vs time comes out to be a straight graph with positive slope.

Considering the question,

A plot of 1/[NOBr] vs time give a straight line with a slope of 2.00 M⁻¹s⁻¹.

<u>Thus, the order of the reaction is 2.</u>

<u>Also, slope is the rate constant of the graph which is :- 2.00 M⁻¹s⁻¹</u>

5 0
3 years ago
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