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2 years ago

I will give u 50 points if u help me

Chemistry

2 answers:

WITCHER [35]2 years ago

7 0

14) gas to liquid to solid
15)chlorine displaces bromine from potassium bromide
16) non-metals
6)vertical
7)similar properties
8)there are patterns
10)decrease
11)very reactive
12) hydrogen
18)lower density than air
19)better insulator
20)metalloids

Fofino [41]2 years ago

6 0

Answer:

a b a

Explanation:

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A hydrate is typically a crystalline compound in which water molecules are chemically bound to another compound or an element.

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Subtract the mass of the CuSO4⋅ 5H2O from the mass of CuSO4 is the right one

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3 years ago

How many grams of MgCl2 would be needed to make 5.3 liters of a 0.25 molar solution? Round your answer to the nearest 0.01,

GuDViN [60]

Answer: 126.1 grams of are needed to make 5.3 L of 0.25 M solution

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

To calculate the mass of solute for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{given mass of solute}}{\text {Molar mass of solute}\times \text{Volume of solution in L}}$

$0.25M=\frac{x}{95.2g/mol\times 5.3L}$

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Thus 126.1 grams of are needed to make 5.3 L of 0.25 M solution

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2 years ago

Why is water a liquid at room temperature instead of a gas?

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Has strong hydrogen bonding which requires alot of energy to overcome hence high boiling point hence liq in room temp

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3 years ago

Read 2 more answers

Your friend's pet hamster weighs 152 grams on Earth. If that hamster traveled to space, what is its weight and why? Answer in co

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3 years ago

what is the pH of a solution that results when 0.010mol HNO3 is added to 500.ml of a solution that is 0.10M in aqueous ammonia a

WARRIOR [948]

Answer : The

pH of a solution is, 8.56

Explanation : Given,

$K_b=1.8\times 10^{-5}$

Concentration of ammonia (base) = 0.10 M

Concentration of ammonium nitrate (salt) = 0.55 M

First we have to calculate the value of $pK_b$ .

The expression used for the calculation of $pK_b$ is,

$pK_b=-\log (K_b)$

Now put the value of $K_b$ in this expression, we get:

$pK_b=-\log (1.8\times 10^{-5})$

$pK_b=5-\log (1.8)$

$pK_b=4.7$

Now we have to calculate the pOH of buffer.

Using Henderson Hesselbach equation :

$pOH=pK_b+\log \frac{[Salt]}{[Base]}$

Now put all the given values in this expression, we get:

$pOH=4.7+\log (\frac{0.55}{0.10})$

$pOH=5.44$

The pOH of buffer is 5.44

Now we have to calculate the pH of a solution.

$pH+pOH=14\\\\pH+5.44=14\\\\pH=14-5.44\\\\pH=8.56$

Thus, the pH of a solution is, 8.56

8 0

3 years ago