What causes objects to speed up, slow down, and change directions?

In a butane lighter 9.8g of butane combines with 35.1g of oxygen to form 29.6g carbon dioxide and how many grams of water

german

If the reacted oxygen is 35.1 g.
H2O mass= (35.1+9.8)-29.6
= 19.3g

4 0

3 years ago

A 32.14 gram sample of a hydrate of MnSO4 was heated thoroughly in a porcelain crucible, until its weight remained constant. Aft

amid [387]

Answer:

MnSO₄.7H₂O

Explanation:

To solve this question, we need to convert the mass of the dehydrated MnSO₄. The difference between mass of the hydrate and dehydrated compound is the mass of water. With the mass we can find the moles of water and the formula of the hydrate:

<em>Moles MnSO₄ -Molar mass: 151g/mol-:</em>

17.51g * (1mol / 151g) = 0.116 moles

<em>Moles H₂O -Molar mass: 18g/mol-:</em>

32.14g-17.51g = 14.63g * (1mol / 18g) = 0.813 moles

The ratio of moles MnSO₄: Moles H₂O represent the amount of water molecules in the hydrate:

0.813mol / 0.116mol = 7 molecules of water.

The hydrate formula is:

8 0

2 years ago

What is the mass in 6.34x10^23 molecules of calcium chloride

Liono4ka [1.6K]

Answer : The mass of calcium chloride is, 116.84 grams

Solution : Given,

Molar mass of calcium chloride, $CaCl_2$ = 110.98 g/mole

Number of molecules of calcium chloride = $6.34\times 10^{23}$

As we know that,

1 mole of calcium chloride contains $6.022\times 10^{23}$ molecules of calcium chloride

or,

1 mole of calcium chloride contains 110.98 grams of calcium chloride

Or, we can say that

As, $6.022\times 10^{23}$ molecules of calcium chloride present in 110.98 grams of calcium chloride

So, $6.34\times 10^{23}$ molecules of calcium chloride present in $\frac{6.34\times 10^{23}}{6.022\times 10^{23}}\times 110.98=116.84grams$ of calcium chloride

Therefore, the mass of calcium chloride is, 116.84 grams

6 0

3 years ago

If i have 340mL of a 1.5 M NaBr solution, What will the concentration be for 1000mL?

Gekata [30.6K]

Answer:

0.51M

Explanation:

Given parameters:

Initial volume of NaBr = 340mL

Initial molarity = 1.5M

Final volume = 1000mL

Unknown:

Final molarity = ?

Solution;

This is a dilution problem whereas the concentration of a compound changes from one to another.

In this kind of problem, we must establish that the number of moles still remains the same.

number of moles initially before diluting = number of moles after dilution

Number of moles = Molarity x volume

Let us find the number of moles;

Number of moles = initial volume x initial molarity

Convert mL to dm³;

1000mL = 1dm³

340mL gives $\frac{340}{1000}$ = 0.34dm³

Number of moles = initial volume x initial molarity = 0.34 x 1.5 = 0.51moles

Now to find the new molarity/concentration;

Final molarity = $\frac{number of moles}{Volume}$ = $\frac{0.51}{1}$ = 0.51M

We can see a massive drop in molarity this is due to dilution of the initial concentration.

6 0

3 years ago

Water has a specific heat of 4.186 J/g*C. How much energy would be required to raise the temperature of 10 g of water by 10 C?

Sladkaya [172]

Energy(heat) required to raise the temperature of water : 418.6 J

<h3>Further explanation </h3>

Heat can be calculated using the formula:

Q = mc∆T

Q = heat, J

m = mass, g

c = specific heat, joules / g ° C

∆T = temperature difference, ° C / K

Specific heat of water = 4.186 J/g*C.

∆T(raise the temperature) : 10° C

mass = 10 g

Heat required :

$\tt Q=m.c.\Delta T\\\\Q=10\times 4.186\times 10\\\\Q=418.6~J$

8 0

3 years ago