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alekssr [168]
3 years ago
9

Which units could NOT be used to measure the acceleration of an object?

Chemistry
1 answer:
vodka [1.7K]3 years ago
4 0
Kg should not be used to measure the acceleration of an object.
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If Maria winks exactly 5 times every minute and she sleeps
Elina [12.6K]

First, we need to find the amount of winks in an hour, to simplify it so we can answer easier. Since there are 60 minutes in an hour, we can multiply 5 by 60 to find the number of winks in an hour.

5*60=300

Now multiply 300 by 8.

300*8=2,400

She winks exactly 2,400 times per day.

Hope this helps!

4 0
3 years ago
Read 2 more answers
How much heat is required, in calories, to raise the temperature of 57.8 g of silver from
Stella [2.4K]

Answer:

87.3 calories of heat is required.

Explanation:

Heat = mcΔT

m= mass, c = specific heat of silver, T = temperature

H= 57.8 g * 0.057 cal/g°C * ( 43.5 - 17 °C)

H = 57.8 * 0.057 * 26.5

H = 87.3069 cal.

The heat required to raise the temperature of 57.8 g of silver from 17 °C to 43.5 °C is 87.3 calories.

8 0
3 years ago
How many g of MgCO3(s) are needed to make 1.2 L of 1.5 M MgCl2(aq) solution?
maw [93]
Molar mass of MgCO3 is 84.313 g/mol
You can calculate this from data on the periodic table:
Molar mass Mg = 24.305g/mol
molar mass C = 12.011g/mol
molar mass O = 15.999g/mol mass 3 mol = 47.997g
Total = 84.313g/mol

Mass to be used in 1.2L of 1.5M solution = 84.313g * 1.2L * 1.5mol /L = 151.763g
I have not taken significant figures into account
The balanced equation you provide is not necessary in this calculation
4 0
3 years ago
2C2H2(g) + 502(g) → 4C02(g) + 2H20(g) reaction type?
ratelena [41]

Answer:

The answer is combustion.

7 0
2 years ago
How many atoms are there in each of the following ?a.)1.50 mol Na c.) 0.250 mol. Sib.)6.755 mol Pb
tamaranim1 [39]
a) \: n = \frac{N}{NA} \\ 1.5 = \frac{x}{6.02 \times \: {10}^{23} } \\ x = 1.5 \times 6.02 \times {10}^{23} \\ x = 9.03 \times {10}^{23} \: atoms \: of \: Na

b) \: n = \frac{N}{NA} \\ 0.25 = \frac{x}{6.02 \times \: {10}^{23} } \\ x = 0.25 \times 6.02 \times {10}^{23} \\ x = 1.505 \times {10}^{23} \: atoms \: of \: Si \:
c) \: n = \frac{N}{NA} \\ 6.755 = \frac{x}{6.02 \times \: {10}^{23} } \\ x = 6.755 \times 6.02 \times {10}^{23} \\ x = 40,6651 \times {10}^{23} \\ x = 4.07 \times {10}^{24} \: atoms \: of \: Pb \:
5 0
3 years ago
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