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1 year ago

Determine the amplitude

Chemistry

1 answer:

vekshin11 year ago

5 0

Answer:I found out that it is c

Explanation:Find an answer to your question Determine the amplitude or period as requested. Period of y = 4sin 1/3x a. 4 b. Pi/4 c. 4pi/3 d. 6pi.

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Find the mass percentage of water in ZnSO4 · 7H20 PLEASE HELP

netineya [11]

Answer:

Hello, I can help!

Explanation:

Molecular Formula: ZnSO4. 7H2O or H14O11SZn

Molecular Weight: 287.6 g/mol

<h2>Computing molar mass (molar weight)</h2>

Functional groups: D, Ph, Me, Et, Bu, AcAc, For, Ts, Tos, Bz, TMS, tBu, Bzl, Bn, Dmg

Common compound names.

Examples of molar mass computations: NaCl, Ca(OH)2, K4[Fe(CN)6], CuSO4*5H2O, water, nitric acid, potassium permanganate, ethanol, fructose.

-Molecular mass (molecular weight) is the mass of one molecule of a substance and is expressed in the unified atomic mass units (u). (1 u is equal to 1/12 the mass of one atom of carbon-12)

-Molar mass (molar weight) is the mass of one mole of a substance and is expressed in g/mol.

God bless you!

8 0

2 years ago

Solid cesium bromide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell

Oxana [17]

Answer:

$\mathbf {density \ d =4.4845 \ g/cm^3}$

Explanation:

Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.

The density d of CsBr in g/cm³ can be calculated by using the formula:

$\mathtt{ density \ d = \dfrac{z \times molar\ mass \ (M)}{ edge \ length \ (a) \ \times avogadro's \ number \ (N)}}$

where;

z = 1 mole of CsBr

edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm

molar mass of CsBr = 212.81 g/mol

avogadro's number = 6.023 × 10²³

$\mathtt{ density \ d = \dfrac{1 \times 212.81}{(4.287 \times 10^{-8})^3 \times 6.023 \times 10^{23}}}$

$\mathtt{ density \ d = \dfrac{ 212.81}{47.4540533}}$

$\mathbf {density \ d =4.4845 \ g/cm^3}$

5 0

2 years ago

An object has a gravitational potential energy that is 800 Joules. Its

olya-2409 [2.1K]

Answer:

36.28 kg

Explanation:

Given data:

Gravitational potential energy of object = 800 J

Height from ground = 2.245 m

Mass of object = ?

Solution:

Formula:

P.E = mgh

g =acceleration due to gravity = 9.8 m/s²

by putting values,

800 J = m× 9.8 m/s² ×2.25 m

J = kgm²/s²

800 kgm²/s² = m× 22.05 m²/s²

m = 800 kgm²/s² /22.05 m²/s²

m = 36.28 kg

6 0

1 year ago

Calcutale Grxn for the following equation at 25°C: 4KClO3(s) → 3KClO4(s) KCl(s)

Step2247 [10]

Answer:

-133.2 kJ

Explanation:

Let's consider the following balanced equation.

4 KClO₃(s) → 3 KClO₄(s) + KCl(s)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression.

ΔG°rxn = 3 mol × ΔG°f(KClO₄(s)) + 1 mol × ΔG°f(KCl(s)) - 4 mol × ΔG°f(KClO₃(s))

ΔG°rxn = 3 mol × (-303.1 kJ/mol) + 1 mol × (-409.1 kJ/mol) - 4 mol × (-296.3 kJ/mol)

ΔG°rxn = -133.2 kJ

5 0

2 years ago

Consider the following unbalanced chemical equation. C5H12(l) + O2(g) → CO2(g) + H2O(l) If 21.9 grams of pentane (C5H12) are bur

Marina CMI [18]

Answer : The mass of water produced will be 32.78 grams.

Explanation : Given,

Mass of $C_5H_{12}$ = 21.9 g

Molar mass of $C_5H_{12}$ = 72.15 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $C_5H_{12}$ .

$\text{Moles of }C_5H_{12}=\frac{\text{Mass of }C_5H_{12}}{\text{Molar mass of }C_5H_{12}}=\frac{21.9g}{72.15g/mole}=0.3035moles$

Now we have to calculate the moles of $H_2O$ .

The balanced chemical reaction will be,

$C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(l)$

From the balanced reaction we conclude that

As, 1 mole of $C_5H_{12}$ react to give 6 moles of $H_2O$

So, 0.3035 moles of $C_5H_{12}$ react to give $0.3035\times 6=1.821$ moles of $H_2O$

Now we have to calculate the mass of $H_2O$ .

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(1.821mole)\times (18g/mole)=32.78g$

Therefore, the mass of water produced will be 32.78 grams.

3 0

2 years ago