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Rufina [12.5K]
2 years ago
10

PLEASE BE FAST

Chemistry
1 answer:
Reil [10]2 years ago
5 0

Answer:

true is true bro 1 is false and athore is true

You might be interested in
2.
katovenus [111]

Answer:

A. 6atm

Explanation:

Using pressure law equation:

P1/T1 = P2/T2

Where;

P1 = initial pressure (atm)

T1 = initial temperature (K)

P2 = final pressure (atm)

T2 = final temperature (K)

According to this question;

P1 = 3 atm

P2 = ?

T1 = 120K

T2 = 240K

Using P1/T1 = P2/T2

3/120 = P2/240

Cross multiply

240 × 3 = P2 × 120

720 = 120P2

P2 = 720/120

P2 = 6atm

4 0
3 years ago
What was the old percentages of copper to zinc before 1983?
vovikov84 [41]

Answer:

95% Copper and 5% Zinc

Explanation:

Easy stuff read in the chemistry book...

3 0
3 years ago
4. How many weeks are equal to 2.94 x 10 to the power of 6 minutes?
OverLord2011 [107]

Answer:

292 weeks

Explanation:

5 0
3 years ago
La formula quimica de la molecula de agua H20 , si H = 1 gramos y O = 16 gramos . Cual es su composicion porcentual ? 88,88 % de
vesna_86 [32]

Answer:

88,88 % de O y 11,11 % de H

Explanation:

La composición porcentual se define como la masa que hay de cada mol de átomo en 100g. Las moles de agua en 100g son:

<em>Masa molar agua:</em>

2H = 2*1g/mol = 2g/mol

1O = 1*16g/mol = 16g/mol

Masa molar = 2 + 16 = 18g/mol

100g H2O * (1mol / 18g) = 5.556 moles H2O.

Moles de hidrógeno:

5.556 moles H2O * (2mol H / 1mol H2O) = 11.11 moles H

Moles Oxígeno = Moles H2O = 5.556 moles

La masa de hidrógeno es:

11.11mol * (1g/mol) 11.11g H

La masa de oxígeno es:

5.556 mol * (16g / 1mol) = 88.89g O

Así, el porcentaje de O es 88.89% y el de H es 11.11%. La opción correcta es:

<h3>88,88 % de O y 11,11 % de H</h3>
7 0
3 years ago
How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?
meriva

Answer:

\large \boxed{\text{77.4 mL}}

Explanation:

                Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

    V/mL:     249

c/mol·L⁻¹:  0.0443     0.285

1. Calculate the moles of Ba(OH)₂

\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1  mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}

3. Calculate the volume of HCl

V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}

8 0
3 years ago
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