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1 year ago

PLEASE HELP!!!!!!!!!!!!!!!!!!!

Chemistry

1 answer:

Alborosie1 year ago

5 0

Based on temperature changes, the phase changes that occur in water is from a solid at A to a liquid at B, and from liquid a B to gas at C.

<h3>What is change of state?</h3>

Change of state is the process whereby a substance changes its state of matter from one form to another as as result of heat changes.

In the graph shown above, water exists in three states at different points.

Water exists as a solid at A
Water exists as a liquid at B
Water exists as a gas at C.

Therefore, the phase changes that occur in water is from a solid at A to a liquid at B, and from liquid a B to gas at C.

Learn more about change of state at: brainly.com/question/18372554

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The stock concentration of NaOH is 0.5M. 0.5M = 0.500mol/1.0L.

valentinak56 [21]

Answer:a) 0.1 mole. b) 4g. c) 2% d) 196 mL

Explanation: in 200mL , 0.1mole

mw NaOH = 40g/mol —> 4g in 0.1 mole

4g in 200mL so 2g in 100mL

density NaOH = 1g/mL so if 4g in 200 mL, 4mL , 196 mL water

5 0

2 years ago

If 5.0 liters H2 (g) at STP is heated to a temperature of 985, pressure remaining constant, the new volume of the gas will be?

vlada-n [284]

Answer:

$V_2=18 \ L \ H_2$

General Formulas and Concepts:

<u>Chemistry - Gas Laws</u>

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
Charles' Law: $\frac{V_1}{T_1} =\frac{V_2}{T_2}$

Explanation:

<u>Step 1: Define</u>

Initial Volume: 5.0 L H₂ gas

Initial Temp: 273 K

Final Temp: 985 K

Final Volume: ?

<u>Step 2: Solve for new volume</u>

Substitute: $\frac{5 \ L \ H_2}{273 \ K} =\frac{x \ L \ H_2}{985 \ K}$
Cross-multiply: $(5 \ L \ H_2)(985 \ K) = (x \ L \ H_2)(273 \ K)$
Multiply: $4925 \ L \ H_2 \cdot K = 273x \ L \ H_2 \cdot K$
Isolate <em>x</em>: $18.0403 \ L \ H_2 = x$
Rewrite: $x=18.0403 \ L \ H_2$

<u>Step 3: Check</u>

<em>We are given 2 sig figs as the smallest. Follow sig fig rules and round.</em>

<em /> $18.0403 \ L \ H_2 \approx 18 \ L \ H_2$ <em />

5 0

2 years ago

When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2

Arlecino [84]

Answer:

1.62

Explanation:

From the given information:

number of moles of benzamide $=\dfrac{70.4 \ g}{121.14 \ g/mol}$

= 0.58 mole

The molality = $\dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent }$

$= \dfrac{0.58 }{0.85 }$

= 0.6837

Using the formula:

$\mathbf {dT = l \times k_f \times m}$

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

∴

2.7 °C = 1 × kf × 0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

number of moles of NH4Cl = $\dfrac{70.4 \ g}{53.491 \ g /mol}$

= 1.316 mol

The molality = $\dfrac{1.316 \ mol}{0.85 \ kg}$

= 1.5484

Thus;

the above kf value is used in determining the Van't Hoff factor for NH4Cl

i.e.

9.9 = l × 3.949 × 1.5484 m

$l = \dfrac{9.9}{3.949 \times 1.5484 \ m}$

l = 1.62

5 0

2 years ago

How many moles of h2o are produced when using 7 moles of h2

Olenka [21]

1) Write the balaced chemical equation:

H2 + 2O2 → 2H2O

2) Infere the molar ratios:

1 mol H2 : 2 mol of water

3) Make the calculus as the direct proportion relation:

[2 mol H2O] / [1 mol H2] * 7 mol H2 = 14 mol H2

As you see you produce the double number of moles of H2O than number of moles of H2 used.

Answer: 14 moles

6 0

3 years ago

If a reaction occurs, what will be the products of the unbalanced reaction below?Zn(s) + HCl(aq) -->

irakobra [83]

The products will be $ZnCl_2(s) + H_2(g)$