What is the answer to mad of two or more substances together in the same place but not chemically combined

2. Which situation best represents conduction?

ikadub [295]

Answer:

c

Explanation:

touching a hot piece of metal. because the energy or the hotness of the pot is touching your hand which is a direct contact.

8 0

3 years ago

Which list correctly describes the usual order of planets inward toward the sun?

cluponka [151]

D is the answer. N, U, sat, J, mars, e, v, mer

5 0

3 years ago

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The density of a 0.75M NaOH solution is 1.1g/ml. Calculate the mass percent of NaOH in the solution

Slav-nsk [51]

Answer:

2.73%

Explanation:

Molarity of NaOH = 0.75 M

Density of solution = 1.1 g/mL

Mass percent of NaOH = ?

Solution:

Number of moles of NaOH = 0.75 mol from 0.75 mol/L

Mass of NaOH = number of moles × molar mass

Mass of NaOH = 0.75 mol × 40 g/mol

Mass of NaOH = 30 g

Mass of solution:

d = m/v

1.1 g/ mL = m/ 1000 mL

m = 1100 g

Percentage of NaOH:

% NaOH = mass of solute / mass of solution × 100

% NaOH = 30 g/ 1100 g × 100

% NaOH = 2.73%

5 0

3 years ago

A gas mixture consists of 4 kg of O2, 5 kg of N2, and 7 kg of CO2. Determine (a) the mass fraction of each component, (b) the mo

kondor19780726 [428]

Answer:

See explanation.

Explanation:

Hello

(a) In this case, one uses the following formulas, which allow to compute the mass fraction of each component: $\% O_2=\frac{m_{O_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{4kg}{4kg+5kg+7kg}*100\%=25\%O_2\\\% N_2=\frac{m_{N_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{5kg}{4kg+5kg+7kg}*100\%=31.25\%N_2\\\% CO_2=\frac{m_{CO_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{7kg}{4kg+5kg+7kg}*100\%=43.75\%CO_2$

(b) For the mole fractions, it is necessary to find all the components' moles by using their molar mass as shown below:

$n_{O_2}=4kgO_2*\frac{1kmolO_2}{32kgO_2} =0.125kmolO_2\\n_{N_2}=5kgN_2*\frac{1kmolN_2}{28kgN_2} =0.179kmolN_2\\n_{CO_2}=7kgCO_2*\frac{1kmolCO_2}{44kgCO_2} =0.159kmolCO_2$

Now, the mole fractions:

$x_{O_2}=\frac{n_{O_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.125kmolO_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=27\%O_2\\x_{N_2}=\frac{n_{N_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.179kmolN_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=38.7\%N_2\\ x_{CO_2}=\frac{n_{CO_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.159kmolCO_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=34.3\%CO_2$