All categories

3 years ago

A 133.8-gram sample of bronze was 10.3% tin by mass. Determine the total mass of tin in the sample.

1 answer:

6 0

You just need to multiply the total mass by the decimal value of the part that is tin. 133.8*0.103=13.8g (following the rules of significant figures).

You might be interested in

What is the volume in liters of hydrogen gas that would be produced by the reaction of 40.0 g of Al with excess HCl at STP accor

DaniilM [7]

The volume in liters of Hydrogen gas = 49.28 L

<h3>Further explanation </h3>

The reaction equation is the chemical formula of reagents and product substances

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products

Reaction

2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)

mol Al :

$\tt mol=\dfrac{mass}{Ar}=\dfrac{40}{27}=1.481$

mol H₂ :

$\tt \dfrac{3}{2}\times 1.481=2.2$

At STP, 1 mol =22.4 L

so the volume of hydrogen :

$\tt 2.2 \times 22.4=49.28~L$

8 0

2 years ago

Read 2 more answers

PLEASE HELP!!

sp2606 [1]

Inbox me for the answer

6 0

2 years ago

A functional group introduces heteroatoms into a carbon chain to increase

ivanzaharov [21]

Answer:

reactivaty

Explanation:

here you go for the answer

5 0

2 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

2 years ago

Bond length is the distance between the centers of two bonded atoms. on the potential energy curve, the bond length is the inter

Alisiya [41]

Answer:Hence, the bond length in HCl is 125 pm.

Explanation:

Bond length : It is an average distance between the nuclei of two bonded atoms in a molecule.

Also given that bond length is the distance between the centers of two bonded atoms. on the potential energy curve, the bond length is the inter-nuclear distance between the two atoms when the potential energy of the system reaches its lowest value. Beyond this if atoms come closer to each other then their will be repulsion between them.

So, the bond length between the Hydrogen and Chlorine atom in HCl molecule is :

$Bond length=radius _{hydrogen}+radius_{chlorine}=25pm+100pm=125pm$

Hence, the bond length in HCl is 125 pm.

5 0

2 years ago

Read 2 more answers