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9 months ago

10

2a. What is the difference between weight and mass?

1 answer:

Dafna1 [17]9 months ago

3 0

Mass is the amount of matter in the body.

Weight is the measure of force.

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Answer: electronic configuration

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In the nucleus, there are 36 protons and 36 neutrons

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2 years ago

Can you help me answer this plz

Zigmanuir [339]

Answer: In Galileo’s time, what was considered the “center of everything”? The Earth! All of the planets and even the Sun went around “us”. Of course, when Galileo saw the moons of Jupiter passing in front of the planet, and disappearing to show up again, it was clear to him that these moons went around Jupiter like our moon goes ‘round the Earth.

That did not set well with the Beliefs of the day, and that is at least one answer!

Explanation:

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2 years ago

Consider the following reaction between mercury(II) chloride and oxalate ion:

siniylev [52]

Answer : The reaction rate will be, $1.9\times 10^{-4}M/s$

Explanation :

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:

$2HgCl_2(aq)+C_2O_2^{4-}(aq)\rightarrow 2Cl^-(aq)+2CO_2(g)+HgCl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_2^{4-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_2^{4-}$

Expression for rate law for first observation:

$3.2\times 10^{-5}=k(0.164)^a(0.15)^b$ ....(1)

Expression for rate law for second observation:

$2.9\times 10^{-4}=k(0.164)^a(0.45)^b$ ....(2)

Expression for rate law for third observation:

$1.4\times 10^{-4}=k(0.082)^a(0.45)^b$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-5}=k(0.246)^a(0.15)^b$ ....(4)

Dividing 1 from 2, we get:

$\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{k(0.164)^a(0.45)^b}{k(0.164)^a(0.15)^b}\\\\9=3^b\\(3)^2=3^b\\b=2$

Dividing 3 from 2, we get:

$\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{k(0.164)^a(0.45)^b}{k(0.082)^a(0.45)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2$

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

Putting values in above rate law, we get:

$3.2\times 10^{-5}=k(0.164)^1(0.15)^2$

$k=8.7\times 10^{-3}M^{-2}s^{-1}$

Now we have to determine the reaction rate when the concentration of $HgCl_2$ is 0.135 M and that of $C_2O_2^{-4}$ is 0.40 M.

$\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2$

$\text{Rate}=(8.7\times 10^{-3})\times (0.135)^1\times (0.40)^2$

$\text{Rate}=1.9\times 10^{-4}M/s$

Therefore, the reaction rate will be, $1.9\times 10^{-4}M/s$

6 0

3 years ago

During laparoscopic surgery , carbon dioxide gas is used to expand the abdomen to help create a larger working space. If 4.80 L

Studentka2010 [4]

Answer:

5.37 L

Explanation:

To solve this problem we need to use the PV=nRT equation.

First we <u>calculate the amount of CO₂</u>, using the initial given conditions for P, V and T:

P = 785 mmHg ⇒ 785/760 = 1.03 atm

V = 4.80 L

T = 18 °C ⇒ 18 + 273.16 = 291.16 K

1.03 atm * 4.80 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 291.16 K

We <u>solve for n</u>:

n = 0.207 mol

Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).

0.98 atm * V = 0.207 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 310.16 K

And we <u>solve for V</u>:

V = 5.37 L

7 0

3 years ago