At STP, how many moles are in 0.880L of He gas?

Why is it advisable to wear long sleeves when students work in a chemistry lab? to provide warmth near the lab refrigerator to p

mote1985 [20]

It is advisable to wear long sleeve when when a student is working in a chemistry lab so that to protect arms from lab chemicals. when someone enter the chemistry lab to wort or to study should be well prepared with appropriate gears and security measure to avoid injury.

9 0

1 year ago

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A substance has a density of 4.5gmL. What is this in kgL? Report your answer with two significant figures.

Ipatiy [6.2K]

Answer:

4.5 kg/L

Explanation:

Density is 4.5g/mL and it means that in 1 mL of volume, the mass contained is 4.5 g.

Let's make a rule of three

1L = 1000 mL

1 mL has a mass of 4.5 g

1000 mL would have 4500 g

Our new density would be 4500 g/L, but we may convert the g to kg

1 kg / 1000 g . 4500 g = 4.5 kg

In conclusion 4.5 g/mL = 4.5 kg/L

7 0

1 year ago

What is the fraction of the hydrogen atom's mass (11h) that is in the nucleus? the mass of proton is 1.007276 u, and the mass of

ololo11 [35]

Hello there!

To determine the fraction of the hydrogen atom's mass that is in the nucleus, we have to keep in mind that a Hydrogen atom has 1 proton and 1 electron. Protons are in the nucleus while electrons are in electron shells surrounding the nucleus. The mass of the nucleus will be equal to the mass of 1 proton and we can express the fraction as follows:

$Mass Fraction= \frac{mass 1 Proton}{mass H atom} = \frac{1,007276 u}{1,007825}=0,9995$

So, the fraction of the hydrogen atom's mass that is in the nucleus is 0,9995. That means that almost all the mass of this atom is at the nucleus.

Have a nice day!

3 0

1 year ago

II. Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.86 g of magnesium ribbo

Leto [7]

Answer:

$Excess=3.53g$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2Mg(s)+O_2(g) \rightarrow 2MgO(s)$

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

$n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg} =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO$

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

$m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg$

Thus, the mass in excess is:

$Excess=3.86g-0.334g\\\\Excess=3.53g$

Regards!

5 0

1 year ago

For the reaction n2(g) + 2h2(g) â n2h4(l), if the percent yield for this reaction is 77.5%, what is the actual mass of hydrazine

Rudiy27

First calculate the moles of N2 and H2 reacted.

moles N2 = 27.7 g / (28 g/mol) = 0.9893 mol

moles H2 = 4.45 g / (2 g/mol) = 2.225 mol

We can see that N2 is the limiting reactant, therefore we base our calculation from that.

Calculating for mass of N2H4 formed:

mass N2H4 = 0.9893 mol N2 * (1 mole N2H4 / 1 mole N2) * 32 g / mol * 0.775

<span>mass N2H4 = 24.53 grams</span>

7 0

2 years ago