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Pepsi [2]
3 years ago
9

One liter of a certain gas has a mass of 1.25 g at STP. the mass of one mole of this gas is?

Chemistry
1 answer:
Damm [24]3 years ago
6 0

the answer is 1.75 i just took the test

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A package of chocolate instant pudding contains 2640 mg of sodium. how many grams of sodium are in the pudding?
alexgriva [62]

Answer:

2.640

Explanation:

Rounding to the nearest whole number would be 3 if that is needed.

4 0
1 year ago
Sulfur reacts with oxygen and creates two compounds. Compound A contains 1.34 g of sulfur for every 0.86 g of oxygen. Compound B
spayn [35]

Answer:

The mass ratio of oxygen rounded to the nearest whole no. 3 : 2

Explanation:

According to Law of Multiple proportion when two elements combine to make two or more different compounds, the mass ratio of the two element in the first compound, when divided by the mass ratio of the second compound , form a simple whole number ratio.

Compound A contains 1.34 g of sulfur for every 0.86 g of oxygen

        \frac{1.34}{0.86}= 1.5

Compound B contains 11.63 g of sulfur for every 10.49 g of oxygen

      \frac{11.643}{10.49}= 1.0

Ratio of oxygen in each compound

   always put the larger number over the smaller number.

\frac{CompoundA}{Compound B}=\frac{1.5}{1.0}=\frac{3}{2}

3 0
3 years ago
Which element does this Bohr model represent? (look at a periodic table if needed)
Serjik [45]
Circulating round the nucleus are the electrons in various orbits of different energy levels. Electrons are negatively charged and represented by the symbol 'e'. In the given image the number of protons are -6. Hence the element in question is Carbon as Carbon has the atomic number 6.
6 0
3 years ago
PLEASE NO LINKS, GIVING BRAINLIEST, 18 POINTS , THANKS AND 5 STAR!!!!!!!!
Korvikt [17]

Answer:

2nd one In my opinion....

5 0
3 years ago
Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen
uranmaximum [27]

Answer:

148.04 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

NO(g) + 1/2 O₂(g) → NO₂(g)      ΔH°rxn = -114.14 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.

ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))

ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol

ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol

ΔH°f(NO(g)) = 148.04 kJ/mol

8 0
3 years ago
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