A(n)____ is a ripened ovary

What happens when you mix water and sugar

alexandr1967 [171]

Answer:

Mixing water and sugar produces a mixture that is called a solution. In chemistry labs, this is often an experiment used to demonstrate the solubility of a solute in a solvent. In this experiment, the solute is sugar, and water is the solvent.

5 0

3 years ago

An element had five valence electrons for bonding. This element is most likely which of the following?

hichkok12 [17]

Atoms in the nitrogen family.
or phosphorus arsenic antimony and bismuth

Have a great day young scientist

3 0

3 years ago

1.00 kg of ice at -10 °C is heated using a Bunsen burner flame until all the ice melts and the temperature reaches 95 °C. A) How

BaLLatris [955]

Answer : The energy required is, 574.2055 KJ

Solution :

The conversions involved in this process are :

$(1):H_2O(s)(-10^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(95^oC)$

Now we have to calculate the enthalpy change or energy.

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = energy required = ?

m = mass of ice = 1 kg = 1000 g

$c_{p,s}$ = specific heat of solid water = $2.09J/g^oC$

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

n = number of moles of ice = $\frac{\text{Mass of ice}}{\text{Molar mass of ice}}=\frac{1000g}{18g/mole}=55.55mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[1000g\times 4.18J/gK\times (0-(-10))^oC]+55.55mole\times 6010J/mole+[1000g\times 2.09J/gK\times (95-0)^oC]$

$\Delta H=574205.5J=574.2055kJ$ (1 KJ = 1000 J)

Therefore, the energy required is, 574.2055 KJ

3 0

3 years ago

Pls help also this one too

nexus9112 [7]

Ok because oreage squalls room inside they're not a fan and I have been so long overdue

6 0

3 years ago

The following reaction is spontaneous as written when the components are in their standard states:

Radda [10]

Answer:

$\large \boxed{\text{1 mol/L}}$

Explanation:

We must use the Nernst equation

E_{\text{cell}} = E_{\text{cell}}^{\circ} - \dfrac{RT}{zF}\ln Q

We want Ecell < 0 for a reverse reaction, so assume it is 0.

Step 1. Calculate E°cell

Anode: 3Zn ⟶ 3Zn²⁺(4 mol·L⁻¹) + 6e⁻; E° = +0.7618 V

<u>Cathode: 2Cr³⁺ (x mol·L⁻¹) + 6e⁻ ⟶ 2Cr; </u> E° = <u>-0.744 V</u>

Overall: 3Zn + 2Cr³⁺(x mol·L⁻¹) ⟶ 3Zn²⁺(4 mol·L⁻¹) + 2Cr; E° = +0.018 V

Step 2. Calculate Q

$\begin{array}{rcl}0 & = & 0.018 - \dfrac{8.314\times 298}{6 \times 96 485} \ln Q\\\\-0.18& = & -0.00428 \ln Q\\\ln Q & = & 4.16\\Q & = & e^{4.16}\\ & = & \mathbf{64.0}\\\end{array}$

3. Calculate [Cr³⁺]

$\begin{array}{rcl}Q & = & \dfrac{\text{[Zn$^{2+}$]$^{3}$}}{\text{[Cr}^{3+}]^{2}}\\\\64.0 & = & \dfrac{4^{3}}{\text{[Cr}^{3+}]^{2}}\\\\\text{[Cr}^{3+}]^{2}& = & \dfrac{64}{64.0}\\\\& = & 1\\\text{[Cr}^{3+}] & = & \textbf{1 mol/L}\\\end{array}\\\text{[Cr$^{3+}$] must be less than $\large \boxed{\textbf{1 mol/L}}$ for the reaction to be spontaneous in the reverse}\\\text{direction.}$

8 0

2 years ago