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Liula [17]
3 years ago
15

The work is in the photo can i get answers please (;´༎ຶٹ༎ຶ`)​

Chemistry
1 answer:
siniylev [52]3 years ago
3 0

Answer:

yes

Explanation:

it is

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A slice of Swiss cheese contains 47 mg of sodium. (a) What is this mass in grams? .047 (b) What is this mass in ounces? (16 oz =
MAVERICK [17]

Answer:

(a) 0.047 g (b) 0.0016 oz (c) 0.0001 lb

Explanation:

The given mass of the sodium in the slice = 47 mg

(a) Mass has to be calculated in grams

The conversion of mg to g is shown below as:

1 mg = 10⁻³ g

So,

<u>Mass of sodium = 47 × 10⁻³ g = 0.047 g</u>

(b) Mass has to be calculated in ounces

The conversion of ounces to g is shown below as:

453.6 g = 16 oz

Or,

1 g = 16 / 453.6 oz

So,

<u>Mass of sodium = (0.047 × 16) / 453.6 oz = 0.0016 oz</u>

(c) Mass has to be calculated in pounds

The conversion of pounds to g is shown below as:

1 lb = 453.6 g

Or,

1 g = 1/ 453.6 lb

So,

<u>Mass of sodium = (0.047 × 1) / 453.6 oz = 0.0001 lb</u>

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3 years ago
Which of the following is a producer?
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C a giraffe that eats the leaves off trees
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Mixing vinegar and with baking soda produces carbon dioxide and water. Is this a chemical change or a physical change?
Sergeeva-Olga [200]
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Mass=1.01 kg; volume =1000cm3
guajiro [1.7K]

Density = 1.01 g/cm^3 or 1.01 kg/dm^3 or 1010 kg/m^3

Density = mass/volume = 1010 g/1000 cm^3 = 1.01 g/cm^3 = 1.01 kg/dm^3

= 1010 kg/m^3


7 0
3 years ago
A 76.0-gram piece of metal at 96.0 °C is placed in 120.0 g of water in a calorimeter at 24.5 °C. The final temperature in the ca
Phoenix [80]

The specific heat capacity of the metal given the data from the question is 0.66 J/gºC

<h3>Data obtained from the question</h3>
  • Mass of metal (M) = 76 g
  • Temperature of metal (T) = 96 °C
  • Mass of water (Mᵥᵥ) = 120 g
  • Temperature of water (Tᵥᵥ) = 24.5 °C
  • Equilibrium temperature (Tₑ) = 31 °C
  • Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
  • Specific heat capacity of metal (C) =?

<h3>How to determine the specific heat capacity of the metal</h3>

The specific heat capacity of the sample of the metal can be obtained as follow:

Heat loss = Heat gain

MC(M –Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

76 × C × (96 – 31) = 120 × 4.184 × (31 – 24.5)

C × 4940 = 3263.52

Divide both side by 4940

C = 3263.52 / 4940

C = 0.66 J/gºC

Learn more about heat transfer:

brainly.com/question/6363778

#SPJ1

6 0
2 years ago
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