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icang [17]
2 years ago
14

Sulfur is yellow, solid, nonmetal. what other elements is likely also a nonmetal?

Chemistry
2 answers:
QveST [7]2 years ago
8 0

Answer: Other nonmetal elements are oxygen, neon, fluorine, nitrogen, carbon, boron, chlorine, argon, and hydrogen.

Explanation:

If you look at the periodic table the elements are grouped together. For example the metals are in the middle, the nonmetals are on either sides (noble gases, earth alkaline, reactive nonmetals, and etc.)

Neporo4naja [7]2 years ago
6 0
A few nonmetals are hydrogen oxygen nitrogen
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Assume that a daily diet of 2000 calories (i.e. 8.37 x 106 J) is converted completely to body heat.
slava [35]

Answer:

(a) the mass of the water is 3704 g

(b) the mass of the water is 199, 285.7 g

Explanation:

Given;

Quantity of heat, H= 8.37 x 10⁶ J

Part (a) mass of water (as sweat) need to evaporate to cool that person off

Latent heat of vaporization of water, Lvap. = 2.26 x 10⁶ J/kg

H = m x Lvap.

m = \frac{H}{L._{vap}} =\frac{8.37 * 10^6.J}{2.26*10^6\ \frac{J}{kg}} = 3.704 \ kg

mass in gram ⇒ 3.704 kg x 1000g = 3704 g

Part (b) quantity of water raised from 25.0 °C to 35.0 °C by 8.37 x 10⁶ J

specific heat capacity of water, C, 4200 J/kg.°C

H = mcΔθ

where;

Δθ is the change in temperature = 35 - 25 = 10°C

m =\frac{H}{c* \delta \theta} = \frac{8.37 *10^6}{4200*10} = 199.2857 kg

mass in gram ⇒ 199.2857 kg x 1000 g = 199285.7 g

5 0
3 years ago
What volume (in liters) of a solution contains 0.14 mol of KCl?
oksano4ka [1.4K]

Answer:

\boxed {\boxed {\sf 0.078 \ L }}

Explanation:

We are asked to find the volume of a solution given the moles of solute and molarity.

Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:

molarity= \frac{moles \ of \ solute}{liters \ of \ solution}

We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.

  • moles of solute = 0.14 mol KCl
  • molarity= 1.8 mol KCl/ L
  • liters of solution=x

Substitute these values/variables into the formula.

1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}

We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}

1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl

1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl

Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.

\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}

x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}

The units of moles of potassium chloride cancel.

x= \frac{0.14 }{1.8 L}

x=0.07777777778 \ L

The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.

x \approx 0.078 \ L

There are approximately <u>0.078 liters of solution.</u>

5 0
3 years ago
I need help please ​
Alex17521 [72]

Answer: If you think about it, B. would be the most reasonable answer with the given factors.

4 0
2 years ago
The compound FeCl3 is made of...
notka56 [123]

The compound FeCl3  is made  of  one atom  and three  atoms  of chlorine


    <u><em>explanation</em></u>

  • FeCl3  is made up of two elements.
  • that is iron  with   Fe chemical symbol  and chlorine with Cl  chemical symbol.
  • There are 3  atoms of chlorine  in FeCl3   because the subscript in front of Cl in FeCl3  is 3.
3 0
3 years ago
Read 2 more answers
HELP!!! An experiment was conducted to determine if the intermolecular forces in two liquids affects their boiling points. Each
ExtremeBDS [4]

Answer: it would be b time taken by the glass surface to dry

Explanation:

i had took the test and i got it right

4 0
2 years ago
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