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3 years ago

PLEASE ANSWER AS FAST AS YOU CAN I'M ON A TIME LIMIT...

Chemistry

1 answer:

Nataly_w [17]3 years ago

8 0

Its C IT WILLBE LESS EXPENSIVETO COLLECT DATA AND MAKE OBSERVATIONS ANOUT ELEMENTS

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What is a cold front

Vlada [557]

A cold front is the leading edge of a cooler mass of air, replacing at ground level a warmer mass of air, which lies within a fairly sharp surface trough of low pressure.

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3 years ago

Read 2 more answers

When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas write the balanved equati

babunello [35]

Answer: When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is $MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}$ .

Explanation:

The word equation is given as maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas.

Now, in terms of chemical formulae this reaction equation will be as follows.

$MnO_{2} + HCl \rightarrow H_{2}O + MnCl_{2} + Cl_{2}$

Here, number of atoms on reactant side are as follows.

Mn = 1
O = 2
H = 1
Cl = 1

Number of atoms on product side are as follows.

Mn = 1
O = 1
H = 2
Cl = 4

To balance this equation, multiply HCl by 4 on reactant side and multiply $H_{2}O$ by 2 on product side. Therefore, the equation can be rewritten as follows.

$MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}$

Hence, number of atoms on reactant side are as follows.

Mn = 1
O = 2
H = 4
Cl = 4

Number of atoms on product side are as follows.

Mn = 1
O = 2
H = 4
Cl = 4

Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.

Thus, we can conclude that when maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is $MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}$ .

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3 years ago

• Briefly discuss the cause of errors in the measurements

rewona [7]

(also called Observational Error) is the difference between a measured quantity and its true value. It includes random error

4 0

3 years ago

A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.

NeTakaya

Answer:

solubility of X in water at 17.0 $^{0}\textrm{C}$ is 0.11 g/mL.

Explanation:

Yes, the solubility of X in water at 17.0 $^{0}\textrm{C}$ can be calculated using the information given.

Let's assume solubility of X in water at 17.0 $^{0}\textrm{C}$ is y g/mL

The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.

So, solubility of X in 1 mL of water = y g

Hence, solubility of X in 36.0 mL of water = 36y g

So, 36y = 3.96

or, y = $\frac{3.96}{36}$ = 0.11

Hence solubility of X in water at 17.0 $^{0}\textrm{C}$ is 0.11 g/mL.

3 0

3 years ago

For each given [H3O+] or [OH-], find the corresponding [OH-] or [H3O+] at 25°C.

-BARSIC- [3]

Answer:

a) [H3O+] = 1.00 E-10 M ⇒ [OH-] = 1.0 E-4 M

b) [H3O+] = 1.00 E-4 M ⇒ [OH-] = 1.0 E-10 M

c) [H3O+] = 9.90 E-6 M ⇒ [OH-] = 1.0 E-9 M

Explanation:

14 = pH + pOH
pH = - Log [H3O+]

a) [H3O+] = 1.00 E-10 M

⇒ pH = - Log(1.00 E-10) = 10

⇒ pOH = 14 - 10 = 4

⇒ 4 = - Log[OH-]

⇒ [OH-] = 1.0 E-4 M

b) [H3O+] = 1.00 E-4 M

⇒ pH = 4

⇒ pOH = 10

⇒ [OH-] = 1.0 E-10 M

c) [H3O+] = 9.90 E-6 M

⇒ pH = 5

⇒ pOH = 9

⇒ [OH-] = 1.0 E-9 M

8 0

3 years ago