All categories

3 years ago

Please help me out i will give you brainlist. 0.500 is wrong

Chemistry

2 answers:

SOVA2 [1]3 years ago

8 0

Answer:

.250

Explanation:

Alik [6]3 years ago

3 0

<h3>Answer: b) 0.250 mol</h3>

============================================

Work Shown:

Using the periodic table, we see that

1 mole of carbon = 12 grams
1 mole of oxygen = 16 grams

These are approximations and these values are often found underneath the atomic symbol. For example, the atomic weight listed under carbon is roughly 12.011 grams. I'm rounding to 2 sig figs in those numbers listed above.

So 1 mole of CO2 is approximately 12+2*16 = 44 grams. The 2 is there since we have 2 oxygens attached to the carbon atom.

-------------------

Since 1 mole of CO2 is 44 grams, we can use that to convert from grams to moles.

11.0 grams of CO2 = (11.0 grams)*(1 mol/44 g) = (11.0/44) mol = 0.250 mol of CO2

In short,

11.0 grams of CO2 = 0.250 mol of CO2

This is approximate.

We don't need to use any of the information in the table.

You might be interested in

2. A reaction vessel is charged with hydrogen iodide, which partially decomposes to molecular hydrogen and iodine:2HI (g) H2(g)

Deffense [45]

Answer:

The value of Kp at this temperature is 7.44*10⁻³

Explanation:

Chemical equilibrium is established when there are two opposite reactions that take place simultaneously at the same speed.

For the general chemical equation for a homogeneous gas phase system:

aA + bB ⇔ cC + dD

where a, b, c and d are the stoichiometric coefficients of compounds A, B, C and D, the equilibrium constant Kp is determined by the following expression:

$Kp=\frac{P_{C} ^{c} *P_{D} ^{d} }{P_{A} ^{a} *P_{B} ^{b} }$

Where Px is the partial pressure of each of the components once equilibrium has been reached and they are expressed in atmospheres. The equilibrium constant Kp depends solely on temperature and is dimensionless.

In the case of the reaction:

2 HI (g) ⇔ H₂ (g) + I₂ (g)

the equilibrium constant Kp is determined by the following expression:

$Kp=\frac{P_{H_{2} } *P_{I_{2} } }{P_{HI} ^{2} }$

The system comes to equilibrium at 425 °C, and

PHI = 0.794 atm
PH2 = 0.0685 atm
PI2 = 0.0685 atm

Replacing:

$Kp=\frac{0.0685*0.0685}{0.794^{2} }$

Kp=7.44*10⁻³

The value of Kp at this temperature is 7.44*10⁻³

4 0

3 years ago

Based on the shapes of the water drops in part D, how does the attractive force between water and wax compare to the attractive

Minchanka [31]

Answer:

Because the cohesive forces inside the droplets are stronger than the adhesive forces between both the drops and the wax, water does not penetrate waxed surfaces. Because the adhesive forces between the liquid and the glass are stronger than the cohesive forces inside the water, water wets glass and spreads out across it.

Explanation:

EDMENTUM

4 0

2 years ago

A compound of a transition metal and iodine is 56.7% metal by mass.How many grams of the metal can be obtained from 630 g of thi

erma4kov [3.2K]

Answer:

357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine

Explanation:

In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.

Given mass of sample compound = 630 g

Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;

56.7 % = 56.7/100 = 0.567

Mass of transition metal = 0.567 * 630 = 357.21 g

Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g

4 0

3 years ago

Nucleic acids are made of which of the following?

abruzzese [7]

Answer: I believe the correct answer would be A.

5 0

2 years ago

Read 2 more answers

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:

$K_1=K_b$

$K_2=\frac{1}{K_w}$

Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

Hence, the value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

7 0

3 years ago