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Hatshy [7]
2 years ago
11

HELP PLEASE!!!!!!! can you help me with science

Chemistry
1 answer:
amid [387]2 years ago
8 0

Answer:

Yes!! What do you need help with ? :)

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How many miles of NaOH will completely react with 10.0 mL of 1.0 M HCl?
Naddik [55]
The reaction between 1 mole of NaOH and 1 mole of HCl creates 1 mole of NaCl and 1 mole of water. Meaning that the moles of HCl needs to equal that of NaOH for the solution to be considered equalized. That being said, you first need to find the numbers miles of HCl by multiplying the volume by the molarity to get 0.01 moles HCl. (1Mx0.01L=0.01). That means that you need 0.01 moles of NaOH. I hope that helps. Let me know if anything is unclear.
6 0
3 years ago
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I need to know 5 and 6
irina [24]
6-cumbustion reaction
7 0
3 years ago
How will adding NaCl affect the freezing point of a solution?
lord [1]

Answer is: adding NaCl will lower the freezing point of a solution.

A solution (in this example solution of sodium chloride) freezes at a lower temperature than does the pure solvent (deionized water).

The higher the solute concentration (sodium chloride), freezing point depression of the solution will be greater.

Equation describing the change in freezing point:  

ΔT = Kf · b · i.

ΔT - temperature change from pure solvent to solution.

Kf - the molal freezing point depression constant.

b -  molality (moles of solute per kilogram of solvent).

i - Van’t Hoff Factor.

Dissociation of sodium chloride in water: NaCl(aq) →  Na⁺(aq) + Cl⁻(aq).

3 0
3 years ago
Read 2 more answers
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0
3 years ago
A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h
Lostsunrise [7]

Answer:

THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K

Explanation:

Mass of alloy = 33 g

Initial temperature of alloy = 93°C

Mass of water = 50 g

Initail temp. of water = 22 °C

Heat capacity of calorimeter = 9.20 J/K

Final temp. = 31.10 °C

specific heat of alloy = unknown

specific heat capacity of water = 4.2 J/g K

Heat = mass * specific heat * change in temperature = m c ΔT

Heat = heat capcity * chage in temperature = Δ H * ΔT

In calorimetry;

Heat lost by the alloy = Heat gained by water + Heat of the calorimeter

                     mc ΔT = mcΔT + Heat capacity * ΔT

33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)

33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1

2042.7 C = 1911 + 83,72

C = 1911 + 83.72 / 2042.7

C = 1994.72 /2042.7

C =0.9765 J/g K

The specific heat of the alloy is 0.9765 J/ g K

5 0
3 years ago
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